Transcript for:
Understanding Exponential and Logarithmic Functions

Mr clown here for clown math today we're going to go through the whole of logic and exponential functions at higher maths this will cover the whole topic including all the past paper questions and transformation of logs and exponential graphs um before we get into that though let's take a little one from our sponsor clar Master sponsor by Le the educational publisher for Scotland they offer viewers of this channel a massive 30% discount just use a discount code clar and maths at leis scotland. co.uk the practice question boots go all the way from level three to Advanced and include National 3 four and five apps as well so you can get any book you need for your studies these are an excellent resource re questions on every topic and contain work Solutions so you can see how the marks are awarded as you go through the book the questions get increasingly difficult so by the time you finish a book you'll be ready for your exams usually 999 just use a desk called clar mask to get 30% off that link down below an exponential function is a function of the form F ofx = 8 VX where a and X are a member of R which is a real number numbers and a is greater than zero so it looks like this F of xal a which we call the base to the power of X which we call the exponent and we say that f ofx is an exponential function to base a okay so that's the basic idea of what an exponential function is so some properties of exponential functions let's take we've got FX = a^ of X when x = 0 f of 0 plug in 0 in s gives you a to^ 0 well anything to^ 0 is 1 so when X is z for any exponential we get one back on the Y AIS in other words it always go through 0 1 similarly if I put one in then F of 1 is just a^ 1 which is a the base so whatever this base is here when X is one we the graph goes through 1 a okay so let's just draw a graph to show you that so here's what an exponential graph looks like if I've got a positive one then I can go Y and X it says it goes through 01 so a Zer up one we just put one on the X a y AIS and it goes through 1 a so along one up a somewhere and it just looks like this it's like a exponential curve as X gets bigger it gets steeper and steeper and steeper and steeper and it kind of approaches zero on this side slowly but it never actually reaches zero let's look at it when we get a negative value so we got y and x and then we've gone through one and it goes through WR one up here so we would get a curve that looks like that approaching Zero from that side and that's when your a is negative okay let's look at some examples of sketching exponential functions example one sketch a curve with equation Y = 6^ of X so we need to work out where it cuts the axis so when x = 0 we get y = 6 0 which is always 1 so we've got 01 it always goes through 01 when it's an exponential unless there's a transformation plus or minus on the end then we work at one other point to work at one other point we just always do x = 1 so y = 6^ 1 which is 6 so we get 1 6 and we get always the same shape so we draw y and x axis just a little curve sketch X and Y note the point1 on the Y AIS and then along one up six we'll just call that v six and it's a curve that goes like this and we're done there e is one of the most important constants in maths and it's often called je number it's an irrational number so it never ends uh can't be written as a fraction and it's approximately 2.718 or it goes on forever the natural exponential function f ofx = e VX is called a natural exponential function it's got a lot of properties for instance when you differentiate it back itself it's got lots of neat things you can do with it we're going to look at graph Transformations for any graph we can reflect it up down left right scale it we've done that before and if youve needed help with graph Transformations look at the functions and graph's topic but each transformation infes a point on the graph and we need to keep track of these so the bestas way to do it is just to make some sort of table to keep track of how things are changing the same rules for transforming exponential graphs have as I've said been studied in the functions and graphs unit so look at that if you need help but here is essentially all the rules you need so very briefly if we add or take away a number on the end we move it up or down so in other words the Y part changes by a if we Times by a or divide by a we scale it it gets bigger by a so we times the Y by a or theide Y by a if we put a minus in front we flip it upside down on the x-axis now this is where it gets tricky when we put it inside the brackets x + a we move to the left and if it's X the same brackets we move to the right that means it's affecting the x coordinate if we get a here then we're compressing it by a units so we half the x-axis value for that if we divide by a here then we need double it there and if we want to reflect over the Y AIS we put a minus X here you're best to see this in examples so example one transformations of exponential functions sketch y = 1 + 2 ^ x so our basic graph is Y = 2 x so let's look at that first so if I just draw a little table we're going to do y = 2x here and we're going to see what happens when it's y = well 1 + 2x so remember from our previous example our basic points you've always got 01 and then you've always got one and then the base is two two now we're looking at what happens when we add one to that well if we add one the Y part changes Z is not changing but we're going to go to two and and one's not changing but we're going to go to three so we get 02 and 1 3 is what we need to draw we're going to get the same shape for cuz there's not been any Reflections so we just draw a y and x axis X and Y and we note that is now 02 and we notice now 1 3 1 3 and we draw basically the same curve it's just a bit steeper than it was it's just moved up by one so we're going to go like this now we could note that here is actually one now cuz remember it never touches zero so now it never touches one cuz it's moved up by one and we're done there example two for transformation of exponential functions in graphs sketch a graph of y = 3 2x so let's look at our basic function which is just going to be y = 3 to VX and then look at what happens when you times that by two there so our first point remember is zero and one and then our other point is one and then Bas is three so what happens when we times the X by two so 0 is the X part 2 * 0 is still 0 so 3^ 0 is 1 so we still get 0 1 no big deal but this time if we times it by 2 when X is 1 then we get 3 to^ 2 3 2 so that becomes 1 n it's been scaled so you get 1 n same shape as as always X and Y we've got 01 and we've got a long one up n so it's just a lot steeper 1 n instead of 13 the ASM toope where would cut what will still never touch the x axis so it just goes up like this and we done there okay so in the previous examples we were given the exponential functions however sometimes we're given information about the function without knowing what it is in this case we use information to Det the equation so other words it's going to be unknown like an A or B or something like that let's do an example where you're given a picture part of a graph of y = a^ x + B is shown find A and B so we need to look at our points to see what they're given us the points they give us are 02 and they give us 39 so I'll just take a note of way points now remember first number is an X and the second number are y so we can just Sub in so when x equals 0 we get a to the power of 0 + b = 2 well a to^ 0 is 1 so 1 + b = 2 solving that b = 1 we've got our B straight away and now for the other point we've got 39 so when x = 3 we get a to power of 3 + b = y is 9 well that's a cubed but B is 1 cuz I just what it out = 9 9 - 1 is 8 a cub = 8 so something * itself * itself = 8 is 2 so a is two we what out our a and our B okay let's move on to graphs of logarithmic functions this time a logarithmic function is a function of the form f of x is log base a x where a and is greater than zero and so is X now there's a relationship between exponentials and logs and that is that if we've got y = a^ x we can rewrite it as log a y = x a simple example would be let's say we had 10^2 = 100 well we can rewrite that as a log because we've got log now base 10 over 100 equal 2 and I always remember these two you can think about this if I've got 10 in the base 10^ squar on the other side equals 100 or conversely when I've got 10 s = 100 the base becomes 10 the answer becomes here and then two the power becomes over that side is the power so let's evaluate some lmic Expressions write 5 Cub = 125 in logarithmic form so we' got 5 cubed = 125 so that means that we now have' got log base 5 of 1 25 right hand side equals our power of three just double check CU I remember this way 5 cubed = 125 so I'm done there okay we have to evaluate log base 4 of 16 so that means that log base 4 of 16 equals some number let's say it's a so rewriting that as an exponential 4 to the power of a = 16 that's how I remember it 4 to the^ of a = 16 so 4 to^ what is 16 2 a = 2 4 2 is 6 okay let's look at some ones like this log a base a 1 what's log base a of 1 well that means we've got log base a 1 equals some number B so that means that a to the power of b = 1 or to get an answer of one when you do a power that means that the power has to be zero so that means b equal 0 so log base a of 1 equals 0 but log of any base of one is always zero that's a big key point there okay what about log base a of a how could we want that out well that means that if I make that equal to some number B A to the power of b equals a I need to get a back so that means if I think of this power is one here B must be one so log base a of a a log of itself same base is always one that's another key point to know okay some basic logarithmic equations example one solve log base 5x = 3 so I just translate that into 5 cubed = x well that's it 5 * 5 25 * 5 is 125 and we're done there log base X of 81 = 2 that means that x^2 = 81 the square root of x something squared is 81 so that means that x = 9 the square Ro of 81 is 9 now notice here we don't get a negative because logs are not defined for negatives and a square root so it's just nine the log function is the inverse of the exponential function so we can reflect the exponential graph in the line Y = X to get the inverse function and that means that when we do our points when x = 1 we get log base a of 1 is zero the points flip to 1 Z instead of 0 1 is it what it was before when it was an exponential and similarly when it was an exponential was 1 a it goes to A1 every logiic graph of the form log base a x passes through 1 0 and A1 so let's have a look at some graphs let's look at log base 6 of X so when x = 1 we get y = log base 6 of 1 our log of 1 is always zero so we get the 0 1 0 and then the other point we need to look at is the base which is six so we look at six we get y = log base 6 of 6 which is 1 so the point is 61 and we get our curve sketch Y and X kind it comes up from here and it goes through 1 Z which is one there and along six up one let's just call that sex one and it just kind of Curves up like that getting slow and slow as it goes and it never touches the y- AIS we save it that's an Asm toope never ever touches that y- AIS just like the exponential if we reflect that never touches the x-axis and it goes up like that so you can see the line Y X is a mirror line because they're inverses so again we can look at graph Transformations for logarithmic graphs and we can scale it move it up down left right and all sorts so just the same rules as functions so functions and graphs check that out if you need to but here's the rules again we did it earli in the video but just there for you in case you need to look at them but this is a topic where practice just makes perfect okay so let's just move straight in an example so transformations of logarithmic functions example one with to sketch a graph of y = log base 5 of x - 3 so again I'll set my table up and this time I'm going to look at y equals the basic function of log base 5 of X and look at those points and then see what happens when it becomes x - 3 here instead so our basic points remember are always 1 Z and then our base which is five and that's 1 you can do a little bit of sum here x - 3 = 1 so adding three to get four so we must get four for X to get back zero and similarly we want x - 3 to be 5 so that gives me 8 here to get back one so we've got our basic logic graphs though so we've got 4 Z so along 40 and we've got an long 81 let's just call that 81 so basically it's a shift to the right by 3 because this is stuck this is basically f ofx - 3 so instead of it not touching the y- axis it'll actually not touch a number three so I'll put a three here just to show that and then just draw a curve I want done there okay transformation of log functions example two sketch the graph of y = log Bas 52 x + 4 so I'm going to start looking at log base 5 of x to start with looking at our points for that so always one zero and then it's going to be a long five because the Bas is five up one so now looking at we need to SP this up into two bits y = log base 5 of 2x is what we'll look at to start with and then we'll add the four so log base 5 of 2x then we'll add the four so let's just do it in pieces so we don't make a mistake so 2x means that must be a half cuz two halfes is one and then we get zero St and then that must be 5ves plus 2 * 5ves is 5 I'm just dividing by two instead of timesing by two you see the Opposites coming into here and we get one still and now we need to add four now we're not adding four to X we're add four to Y so that becomes a half four up the Y AIS 5ves and five so we've got bit scaling going on here but you'll still get the same sort of graph so if I draw the y and x axis first of all I'll show you it by drawing it separately so let's draw our normal log base 5x is this one it goes along 1 Z then somewhere 51 so then we can draw log base 52x if we have to well all that's happen with that one is the points have changed it's got a half zero so that's a half and we've got 5 halfes 1 so we'll just take a note 5 halves 2.5 and 1 that's still basically the same graph nothing else is really changed and then we've moved it up by four for the final part so our final answer over all you can just jump straight in with final answer if you're happy with that we've got a half four and we've got 2.55 and again it will still hug the Y AIS go something like that and that is our final answer y = log base 5 of 2x + 4 and we're done there again we can look at logiic equations with messing constants and look at graph to see if we can work it those messing constants as we did in the exponential one so let's get an EV that straight away with an example missing constants example one the graph of y = log base a of X is shown find a well 1 Z is never going to help because it's always 1 Zer so I look at 51 so looking at 51 that means X and Y so we've got Y is 1 = log base a of 5 so that means that a to^ 1 = 5 or our on a is 5 and we're done there here's another example of missing constants y = log base M of x + K is shown from K and M well we're given two points we sub them in so let's look at- 40 so we got y = 0 of log base m of -4 + K that means that Translating that to an exponential M to the^ of 0 = - 4 + K the m^ 0 rememberers 1 is - 4 + K so 5 = K or you could have realized that it's moved Along by five to the left because it usually is at one if you want to think it like that that's a nice way to do it as well and then looking at our 51 well we've got Y which is 1 equals log base M of 5 plus K well we now know K is also five so that means M to the^ of 1 = 10 so M = 10 and what done there the base is 10 okay let's start looking at some rules of logarithmic functions away from graphs and just quick reminder logarithmic functions function of form log base a of X where a and X are greater than zero and remember the relationship if you've got y = a^ x is an exponential it can get transferred into a log by log Bas a of y = x a couple of rules also remember log base a of 1 = 0 and log base a of a log of itself is always one so log of one is always zero log of itself is always one let's have a look at some tasks we have to use our calculator to write the three de place the value of the following so I'm going to do this for you just to get an idea of logs so log base 10 of 10 is well one and let's get our calculator out so log base 10 in a calculator and my one is just called the log button and yours probably as well so you just press log and two and it says 0.301 10 or5 don't get confused here it's not two 10 power of something is five it is 0.699 don't about my calculator doing something else in there it's just the final answer we need log of 800 is 2.93 if I want log base 10 ofun 10 I press log base 10 s < TK 10 and make sure I close my brackets I suppose and I get 0.5 I don't know if that is surprising to you but it shouldn't be 10 to power of something Asun 10 where something must be power half 10 power half is the same as root 10 and then last one log of 0.6 is- 0.222 so that's how you use a calculator to work out a log as a number okay now this is the second task I want to look at I want you to look to see if there a relationship between timesing logs and adding logs so we've got some numbers here and we have to work at log 10 a log 10 a * log 10 B log 10 a plus log 10 B now you might know you know the answer because You' been taught it but for the sake of this video I'll show you anyway so I've got log 10 2 fre is 6 so let's look at log 6 we get 0.778 47 is 28 so log 28 is 1.44 7 and 8 sixes is 48 so that gives me 1. 681 so now we want to do log base e * log base B well that means I'm doing log 2 * log 3 0.144 now words not related to log a * B let's do it for all of them 4 S is 28 so log 4 * log 7 0.509 again no relationship it would seem log 8 * log 6 so we basically Pro 0.703 that log of a does not equal log a * log B but does it equal log a plus log B well you probably know the answer of this so I've got log 2 plus log 3 look at Ag 0.778 and if you do it for all of them you'll get the same answer 1.44 7 and you'll get 1. 681 now what go for logs log a = log a plus log B same base so it has to be the same base so it's a it's an extension of the laws of exponents because exponents logs are related and that's where it comes from so check an example log 2 base 8 plus log base 2 of 4 would equal log base 2 of 8 * 4 which is 32 now can we work that out that means that 2 to the power of something I'll just call it a = 32 2 2 is 4 * 2 is 8 * 2 is 16 * 2 is 32 e = 5 so log base 2 of 32 = 5 a few checking examples just making sure we're happy with that rule we'll just leave our answer is a log if we have to if we can't work it out log base 5 3 + log base 5 6 is log base 5 3 * 6 is 18 can't find a power of five it makes 18 so I just stop log base 93 plus log base 9 27 so that's log base 9 3 * 27 so that's 81 but I can work that out 9 squar is 81 so that's two and we're done there now I'm not going to do the next one it follows on if log a b = log a plus log B log a over log log A over B is log a minus log B so there's a gole there if we're dividing A and B in a log you can just take away two logs so so for instance log base 2 of 8us log base 2 of 4 is just equal to log base 2 of a over 4 of that's log base 2 of 8/ 4 is 2 log of itself is 1 so that gave me one couple of other examples using that rule so if to combine log base 7 of 15 minus log Bas 7 of 3 so I can use the division rule log base 7 15 over 3 log base 7 15 / 3 is 5 Second example log base 4 of 6us L log base 4 3 make sure the base the same log base 4 of 6 over 3 that's log base 4 of 2 so that means 4 to the power of something let's call it a equal 2 well we could say is a half then Square Ro of 4 is two so it's a half and we're done there another rules of logs that we can show quite similarly is that log base of anything of a to^ B I can just bring the B down to front it equals B log base X of a so in other words log base 2 of 2 cubed since it's all power I can take the power over the front and that just leaves 2 two but remember log base itself is 1 so that's 3 * 1 which is three let's have a look at some questions using that rule Express two log base 73 and log base 7 x so that means I'm just I've got a two at the front so I can take the three and then make it squared so that becomes log base 7 of 9 simplify 2 log base 9 3 well that's just log base 9 of 3 S that's log base 9 of 9 1 okay so there's a note of all of our rules of logs log a b = log a plus b log A over B is log a minus B log a to the power of B is B log a and then remember log of one is zero and log of itself is one and we can combine these in much more tricky questions so let's look at combining rules of logs example one evaluate log base 12 10 + log base 12 6 minus log base 125 so rules of logs I can add means times so that's 10 * 6 which is 60 minus base 12 of 5 minus means divide 60 over 5 log base 12 of 12 cuz 5 12 is 60 which gives me one and we're done here example two log base 6 of 4 + 2 log base 6 of 3 so I can't add yet because that two is in the way so I'll leave it there and move it up to a power 3^ s so that's log base 6 of 4 log base 6 of 9 and now add means Times log B 6 of 49 is 36 so I'm look for six to the power of something is 36 that's two 6^ s is 36 okay let I'm look at some further logarithmic equations we previously did the ones which were basic log base a of x = 2 or something like that let's look at combining the rules of logs so example One log base a 13 plus log base a x = log base a 273 looks quite complicated but we can just combine the left a and we've got 13x 1 log = 1 log now we've got a log on both sides here if you've got a log on both sides and it's the same base and that's all you're left with you can just eliminate the logs cancel each other out so that means that 13x = 273 and then we can just divide by 13 so that means that xals use a calculator if you want 21 13 21s is 273 and we're done there so example two for the logarithmic equations log base 11 of 4x + 3 - log base 11 of 2x - 3 equal 1 and it tells you that X is greater than 3 halves so we can combine the left cuz 11 is the same base and a 4X + 3 over cuz it's- 2x - - 3 = 1 now so we can just eliminate the logs here cuz there's no there only a log on one side but we can eliminate it because that means remember remember log base 10 over 100 = 2 means 10^ 2 = 100 that to^ that = 100 so that to power of that equals all this so we can just save it 4x + 3 over 2x - 3 = 11^ of 1 T in both sides by 2x - 3 4x + 3 is 11 2x - 3 a little equation to solve expand the right hand side to get 22x - 33 move all the X's to the right so 22 - 4 that makes 18x and move the- 33 over + 33 + 3 is + 36 dividing through by 18 x is 36 over 18 which is 2 and we're done there okay for the logarithmic equations example three log base a of 2 p + 1 plus log Bas a of 3 p - 10 = log base a of 11 P for p is greater than 4 okay let's solve this don't worry about it's a log base a or anything let's just see what happens when we combine it it's all the same base so we've got log base a 2 p + 1 3 p - 10 cuz ADD means times was log base a of 11 P so notice we've got a log of something equals a log of something same base so log eliminate it doesn't matter if it's a base a cuz you're never going to work it out anyway so that means that 2 p + 1 3 p - 10 equal 11 P well that's double brackets equals to something so I need to get that equal to zero so I'm going to have to expand the brackets so 2 * 3 is 6 p^ 2 - 20 p + 3 p - 10 you can use foil or a good or anything else you want moving everything to the left you get 6 p^ 2 - 20 p + 3 p - 11 P - 10 is 0 so now I need to get in p^ s p number so I've got - 20 plus 3 is -7 -7 - 11 is - 28 P - 10 now luckily there's a common factor because six is a nasty number common factor is two I think 3 p ^ 2 - 14 P - 5 = 0 and then it's quadratic formula or hopefully double brackets so that gives me two double brackets 0o so what we got 3 p and p 3 5 is 15 minus 1 is 14 so be there will be it and you've got 3 p + 1 and it will be - 5 just double check 3 p ^ 2 - 15 p + 1 is -4 P - 5 yep so that means the first bracket equals z or the second bracket equals z so that means that P = minus A3 or P = 5 initial conditions are important here p is always bigger than four so that means that we can just eliminate that one and therefore P = 5 since p is four and we done there okay example four for for logarithmic equations solve log base 2 7 = log base 2x plus a number three okay this looks more complicated than it actually is this number three is going to bother a lot of people cuz I can't get rid of the logs or anything but was a trick and the trick is this remember log base of itself equals 1 so if I want to introduce a log here I can just say 3 * 1 and just make it a special One log base 2 two and I pick base two because everything else is base two and then I can combine the logs on the right so it's a little trick if it is worth knowing log base 2x plus log base 2 2 cubed combining the logs on the right hand side we get log base 2 * so it's 2 Cub is 8 8 x now logs can just disappear I've got a log of something equals a log of something same base so that means it's 7 = 8X so x = 78 and we're done there okay remember we talked about e on again this is what e is and you can write that as an equation by the way as 1 + 1/ n the^ n as n goes to infinity and if you substitute VAR L values of n you will get a value very close to e so it's basically the limit as n goes to Infinity yeah need to know that or work with that but it's worth just seeing but there is e 2.71 and we're going to look at exponentials and logs to base e Now log base e ex is a bit clunky and since we use base e over the time we give it a special notation RN because it's called the natural logarithm of X so we use RN X to mean log base e of X log base e of e would also equal one so now e l e equal 1 we without a calculator we have to write down the values of log base e of e s so that means that we've got two Ln e which is just two Ln e or log base e of e is one remember we've got five RN e we've got five root e we're going to have to fix a little bit so we've got RN for log Bas e and then root e is e of a half well that's just a half of l e which is just a half it's the same as the one above Ln cube root of e well that's Ln e of a third which is a third a third Ln e which is 1/3 and this one that looks a little bit tricky but we've got 1 over e of a thir which is e minus a thir which is minus a thir RN which is minus a thir L zero cannot be found easy way to explain why on this one as we're looking at rn0 that means the same me as log base e of Z well let's say that a that means we're saying that e to the power of a equals z a number which is 2 something to the power of another number equals zero is impossible and now way look at it as a graph if you think of a graph always goes through one and then 1 a and it never touches the xaxis in other words never xals 0 is undefined so there you are okay with to use a calculator to work out log base e of a and solve RN x = 9 log base e of a we for the RN button press a 2 point 079 very simple solve Ln xal 9 so you might like to think of that as log base e of x = 9 and that means e to^ of 9 = x so I do e to^ 9 in my calculator so to get e i press shift and log on my calculator and put in the number 9 so X = 81030 say and we're done there okay to simplify this Express answer the form a plus log b e b log base e c so you need to be careful with this let's have a look at 4 log base e 2 eus 3 log Bas E2 three so we can put some powers in log base e 2 e to 4us log base e 3 e cubed let's work out those Powers then so we've got log base e of 2 2 is 4 * 2 is 8 * 2 is 16 e to 4 minus log base e of 27 e cubed now looking at the form that we want it in we want a number plus log base e minus log Bas e of C so I'm going to have to get a number somehow so I'm just going to split this up into two logs log base e of 16 * E4 so plus log base e of e to the 4 and then do the same for the take away one so I've got log base e of 27 plus log base e of e cubed so that equals log base e of 16 plus just think of a 4 coming down one 4 log base e of e well that's just 4 minus log base e of 27 and again log base e of e where a power of three means the three can jump to the front so it's just plus three and now we can get our numbers coming out we've got two logs taken away 4 - 3 is 1 or to write it in the form we wanted 1 + log base e16 - log Bas e27 quite tricky that one to get it in the correct form solving exponential equations with unknown exponents so now what's with an X in the power position example one is solve e to^ x = 7 now the way you solve an exponential equation is you take the log of both sides because you need to get rid of that power in the power position because you can't think in terms of powers so you take logs and natural logs if it's base e if it's any other base like 10 you'll take log base 10 if it's two log base 2 but since it's natural log so we can just say Ln e VX = ln7 to keep it balanced and then X comes to the front we've got Ln e equal ln7 but remember Ln e is always one so that that means that X = ln7 and with done there we could just work that out using a calculator ln7 is 1.94 6 and we're done there let's look at another few of solving equations with unknown exponents so I've got a bunch here to try e to^ x is 10 so we say the Ln e x = Ln 10 so that means that X Ln e equal Ln 10 so X is ln1 and you can work that out as many deal places you want using a calculator ln1 is 2.32 303 okay another one e x is a 1,000 so Ln e VX is Ln a th000 so X Ln e equal Ln a000 let go finally x equals Ln one remember Ln a000 let's see what that is 6.98 another one 2 e VX equal 0.3 so again pick the log of both sides Ln or of 2 e VX = Ln of 0.3 so now we can split that up into two logs log base 2 plus log base X so now we can split that up into two logs so we're going to have rn2 plus RN e x = Ln 0.3 Ln 2 + x Ln e = Ln 0.3 so finally remember this x Ln e just X so we've got Ln 2 + x = Ln 0.3 so X is Ln 0.3 minus ln2 you can just use a calculator to work that out or you can combine it Ln 0.3 over two minus 1. 897 looking at D we've got e to x / 2 = 5 this is D sorry so we can take the log of both sides take a power to the front then to get RN out so that means that X over 2 = Log 5 so X is 2 Log 5 or log 5^ s if you prefer doesn't really matter if you're going to use a calculator 2 Log 5 is 3219 question e e 2x = 18 so log e 2x = log 18 2x comes in front to get log base e of log 18 so 2x = log 18 so X is log 18 all / 2 or half of log 18 or 18 power over half we just keep it like that it's much simpler log 18 all divided by 2 is 1.44 and then the last one e- x/ 2 = one so log both sides take Min - x/ 2 the front that gives me X over 2 or minus equal log of one well what's the log of one that's what you're asking yourself here log one is zero remember which means that x = 0 and we're done there okay some more solving equations with unknown exponents solve 7^ y = 9 so this time we're not going to take the natural log take any base you want actually it doesn't make any difference whatsoever so this time we got 7 y = 9 so we take the naturual log or you can take log 10 either two options don't pick any other log cuz most of the time calculators will only do e or 10 I'll use 10 this time log base 10 of 7 y just to show you that it's fine so the Y comes to the front and very we divide so getting a calculator to work all of that out log base 10 remember log of 9 divided by make sure you close that bracket log of 7 is 1.29 to three decimal places and we're done there 5 to 3x + 1 = 40 so let's do log base 10 of 5 3x + 1 is log base 10 of 40 so 3x + 1 can come to the front to get log base 10 of 5 = log base 10 of 40 so 3x + 1 = log base 10 of 40 over log base 10 of 5 log 40 / Log 5 is 2292 so 3x + 1 is 2292 so 3x taken away 1 is 1. 292 so finally we get x equals just divide 1. 292 / 3 0.431 and we're done there okay so looking at exponential growth and Decay we PR exponential function are sometimes known as growth and Decay functions because the exponential go or exponential decay and these model real life situations quite often think of accumulation or think of a virus spreading like covid-19 that's exponential growth okay H or radioactive decay so we can look at some equations of exponential growth in Decay and solve some things so imagine this one exponential growth in Decay example one example one says Mass G Gams of a radi sample after the time T years is given by a formula G = 100 e to Theus 3T what is an initial mass of a radioactive substance in the sample and find the half life so initial mass is always initial T equals 0 so part A when t = 0 we get G = 100 e to the- 3 * 0 which is equal to 100 9 * 10 is always a number in front here and now says find the half life of the radioactive substance we know the initial mass is 100 so when halves a half life we need to half 100 so we become 50 so G is equal to 50 so we get 50 = 100 e- 3T 50 over 100 there's that half coming out again is eus 3T so have to solve that equation now notice we could have not KN the initial it could have just been like a and because when you you half it you get a half a here and when you divide it you get a half here essentially what I'm trying to say is if you don't know that number whatever fraction is goes here a half or a percent okay let's continue 12 = eus 3T so taking the log of both sides log of a half natural log CU Bas e is e- 3 t with a log so log a half = - 3 t log e but remember log e = 1 so log a half = -3 T and we just we finished T must be log a half over -3 use a calculator for that bit LA and a half 0.5 /us 3 I'll put that in Brackets actually is 0.231 and what is our units on this one years doesn't ask us to change it to um months or anything so I'm done there indicate example two the world population in billions T years after 1950 is given by P = 2.54 e over 0.178 T what was the world population in 1950 so 1950 this is part A sorry t equal 0 so P equals it's going to be 2.54 but I'll show it exactly 2 .54 e to 0.178 * 0 so remember that's 2.54 * e of a 0 e 0 is 1 2.54 billion make sure you answer that question write it in words if you want Part B find to the nearest G of a Time taken for the world population to double so if it doubles we need to double the population so that means that the population p is 2 * 2.54 which is 5.08 so we can say that our sum becomes 5.08 = 2.54 e 0.078 t but remember we're going to divide by that here which is going to give us a two so if we want to just go straight away we could have say that two equals just this and we would have got a bit quicker so dividing by 2.54 we get 5.08 over 2.54 is e of 0.178 T that's two e to the 0.178 t log 2 then is log e 0.01 78t taking the 0.078 T to the front you get log e which is one remember so that means that log 2 = 0.178 T So T is log 2 over 0.178 using a calculator 38941 and that's years ient to answer the question year 39 years is how long it takes for the population to double um and then you could probably say well that would be 1950 6070 89 is when it doubled but we don't need to do that for the now okay let's look at experimental data on learning models so the Rel of an experiment make sure there's an exponential relationship between the Y and ax or any other availables that is so they can be difficult to interpret results cuz rapidly increasing or rapidly decreasing um examples of these graphs would be like bacteria a sample rabid populations energy produced by earthquakes and so on but we can use logs to convert the exponential graphs the linear graphs by converting the axis to logarithmic scales see that in an example but if you do do that you get a straight line let's have a look see we've got polinomial equations of the form let's say ax B so let's say we've got y = ax B now we used to taking logs of both sides to solve these so let's just do that log let's say base 10 of Y is log base 10 over ax B right now we can use our rules of logs to separate that out so log base 10 of Y would equal log base 10 of a * x b so plus log base 10 of X the power of B now that means we've got log base T of Y = log base T of a plus b log base T of X let me Flip It On Us log base T of y equal B log base 10 of X Plus log base T of a if I take away these logs for a minute here that looks like y = mx + C exactly what it is except our axes are log base 10 Y and log base 10 x so we've got an equation y = mx + C if we use a big big y equal MX plus a Big C then essentially Our Big Y equal log base 10 y our big our m is just equal to B and our C or X is log base 10 x and our c is log base 10 a so if we have taken our original graph y it goes ax to B and drawn it normally we would get an exponential curve but if we had now instead change the axis to log y any log but we use base 10 for this and log X we would get a straight line and it's c is where it cuts the Y AIS so that would be log base 10 of a and the gradient would be some B okay so that's what we're going to do when we do exponential data we're going to do examples where we change the axes okay here's an example the results from an experiment are not to below and it's told us log base 10 x v numbers log base 10 y v numbers the relation between these data can be written in the form why it was a x to B find the values of A and B in state Y in terms of X now some teachers will teach you to memorize this stuff but I prefer to stick with starting with y B and then coming up with it myself every single time so I start with Y = ax B and I take the log of both sides log base 10 of Y is log base 10 of ax B and I just use my rules of logs I did before log base 10 of Y = log base 10 of a plus log base 10 of x to the B log base 10 of Y is log base 10 of a + B log base 10 x I want MX plus C remember so log base 10 y = b log base 10 X Plus log base 10 a and now I can see that's like a y = an m and X plus a c so now I can use that knowledge what about our gradient well our gradient can be determined by any points because our data is given to in terms of log X and log y so our gradient is Y2 - y1 / X2 - X1 so working out our gradient let's just pick these two numbers and these two numbers so 1.67 minus 1.33 for our y's and 2.29 - 1.7 for our X's now that simplifies the 0.34 over 0.59 which is 0.58 to 2dp so that means that since m is b b equal 0.58 so we've already got log base 10 of Y = 0.58 log base T of x plus log base 10 of a so let's sub in a point we've got 1.33 and 1.7 so 1.70 = 0.58 * 1.33 plus log base 10 of a now notice I'm sub I'm straight in for here because it's qu Us in terms of log Bas 10 x and a so we can just work on that and solve it 1.70 equals 0.9 986 use a calculator plus log base 10 a taking away V numbers you get 0.344 so log base 10 a equal 0.344 so 10 power is 0.344 equal a using the calculator that gives me 2.28 to three decimal places or 2.2 1 I suppose so a = 2.21 so now we have to write it in terms of Y and X remember it was originally y = a x b so we've done it y = 2.21 x power of 0.58 and we're done there so the first time we did this the X was down below and B was a power we can also do this when X is a power so a little bit different a the power of X but again starting the same way this time we take a log of both sides as usual using the rules of logs you can get log base 10 of a plus log base 10 of B to the X take the X to the front so there you are there's our mx + C so putting that in that order I'll write it as log base 10 B and just put a bracket around the X Plus log base 10 a so you can really see that so that is like y = m x + C that's basically what it is where y = log base 10 y m = log base 10 B well X is just X so since X is just X and C is log base 10 a what we're talking about here is a graph where the xaxis is still X but the Y AIS is logarithmic so when X is in the power that's what happens and then it still cuts at some point on C at log base 10 a but again I wouldn't worry about that because I would just do it like this every single time and you get it you get there anyway so let's look at another example like that was experiment sh Bel but this time it tells us in terms of X and Y not log X and log Y and it says it can be formed in the rati y a b x find the formula for y in terms of X so y = AB VX log what just pick base 10 of Y log base 10 of a v x I would just start for a I would not memorize things okay so log base tan of y equal log base tan of a plus log base tan of b v x log base 10 of Y log base 10 of A+ X log base 10 of B we wrting that in the form y = mx + C log base 10 y equals we want the X bit first so log base 10 B * X Plus log Bas 10 a so that means we've got y = m x + C and we're done there so we now got in correct form so we need to make sure our axes are the same so our y AIS is log 10 Y and our x axis is just the x axis so our origal data is not given in that form it's given in X and Y so I'm going to have to change it to X and log y so that's quite easy I just put X and log y log base 10 will say of Y so original xes were 1.30 2.0 uh 2.3 0 and 2.8 Z so then we just put in now we're just check our wi 7.69 12.94 16.2 and 23.10 so I'll do one but you can do the rest yourself the log of 7.69 so the first number is 0.885 or 0.89 just do it to two decimal places cuz the rest are I then do the next log of Y which is 1.11 the next log was 1.21 and the next one's 1.36 I'm getting these numbers just by taking the log base 10 of this this this and this are yse to make it the same scale so now our gradient we can work out because we've now got the correct way our gradient is let's say Y2 - y1 over X2 - X1 so you can work that out to get 0.22 over 0.7 which is 0.31 to two decimal places but remember our gradient is log 10 B so that means that log 10 b equals 0.31 so 10 of 0.31 equal B calculate for that 2.04 to two decimal places so there's our B so we now need to work out our C so we've got log 10 y = log 10 BX plus log 10 a so let's just pick another Point Let's just pick 1.3 and 0. 89 so log 10 Y is 0.89 = log 10 B Well Log 10 B remember 0.31 don't need to work it out again we know our B but is 0.31 times our X so I picked 0.89 so it's 1.3 plus log 10 a and we just need to work that out so 0.89 equal 0.40 plus log 10 a so log 10 a = 0 49 so that means that 10 the^ 0.49 = a which is 3.09 we' got our a we've got our B remember y = a to the^ of X so y = 3.09 time 2.04 to the^ of X is our relationship and we're done there now you might find that quite tricky best sometimes they actually lead you through this by showing you the graphs and showing you the axes as well have a look at the past paper questions P paper questions for all of this topic logarithmic functions exponential functions are coming up now okay evaluate numerical expressions ask High Master 15 paper one question six evaluate log 62 plus a thir of log 627 so I'll keep the log 612 to start with but then I'll write that as plus of log 6 27 to the power 1/3 because we can take that up so that's log 612 plus is 7 to the^ thir is a cube root of 27 which is 3 so log 6 3 and add in logs means we can just time it in one log and the numbers times together so that's log 6 of 12 * 3 that's log 6 36 which is equal to 2 and that's because remember 6^ 2 = 36 so two is our final answer and we're done there evaluating the expressions for logs s high mass 2016 paper 1 question 14 part A evaluate log 525 2 cuz remember means 5^2 = 25 it's as simple as that part B hands solve log 4x + log 4x - x = log 525 so looking at our left hand side we combine combine that into one log by saying that it's x * x - 6 so I'll do that to start with log 4 x X - 6 and the right hand side log 525 well we know that's two so we can just write two so then we can use our rules of logs to say that as a base to the^ of two equals left hand side so in other words 4^2 = x * x - 6 they eliminate the log now we're just going to get a quadratic 16 = x^2 - 6 x moveing over to the same side you get x^2 - 6 x - 16 = 0 so you've got a quadratic to solve hopefully that's factorizable x and x 8 and 2 - 8 + 2 because - 8 + 2 is - 6 - 8 * 2 is -6 so that gives me x = -2 or x = 8 but X = to-2 is not a valid answer because the log of zero or anything negative is undefined so X has to be greater than six so we can just score that out and say that X has to be greater than six is the answer so our final answer is X = to 8 and we're done there okay evaluate numerical expressions sqa higher maths 2018 paper one question 6 find the value of log 5250 minus a f of log 58 so we're going to combine these so we'll keep the log 5250 hanging out for a minute but we can mean see minus log five 8 to the power 3 because you can take coefficients up as a power we'll work out that 8 to the^ of thir then so that's log 5250 minus remember to the power of f is a cube root so it's log 52 and now we can combine a minus because a minus means you can do this number divided by this number so as one log we get Log 5 250 over 2 that's Log 5 125 so 5 squ is 25 so 5 cubed is 125 so the answer is three because 5 cubed is 125 and we're done there okay evaluate numerical expressions ask high mass 2019 paper 1 question 14 part a valate log 10 4+ 2 log 105 so we've got log 10 4+ log 10 5^ SAR I can take the two up so that's log 10 4 + log 10 25 and then we can combine that because we can times the numbers together so it's log 10 4 * 25 which is log 10 100 log 10 100 is equal to 2 because 10^ 2 = 100 and there we are Part B solve log 2 7X - 2 - log 2 3 = 5 so we'll combine that into one log straight away by writing log to 7 x - 2 all over 3 because a minus means we can divide that equals five so when using our rules of logs to eliminate we've got a log on one side the number so 2 to the power 5 equals this so I can write that 2 to the^ 5 is 7 a x - 2 over 3 so we need to work out 2 to the^ 5 well 2 2 is 4 * 2 is 8 * 2 is 16 * 2 is 32 so we get 32 so 32 = 7 x - 2 over 3 so we can times 32 by 3 32 * 3 is 96 so 96 is 7 x - 2 add two to both sides so 98 is equal to 7 7 x dividing through by 7 then you get 14 = X or nicer x = 14 and we done there SQ High m 2022 paper 1 question two evaluate to log 3x minus log 34 so it's just just a thre mark question this one the first one we're going to apply is the idea that 2 log 36 can be simplify to log 36 s so you get your first Mark for simplifying this first log to log base 3 of 6^ squar and make it still WR minus log base 3 of 4 as long as you wrote this there's your first Mark there and then our second mark for combining the takeaways into one log as a divide in other words log base 3 6^ 2 over 4 there's our second Mark there let's just take a note of that and of course our third Mark for an evaluating what that is so working it out so let's do some work on that log base 3 36 over 4al = log base 3 of 9 well 3 S is 9 so log base 3 of 9 = 2 and there is our final Mark right at the end for getting two just a little note on this if you did the correct answer but did no working on this question you would get no marks you have to show your steps to show that you understood it otherwise just think you were guessing on this question hi M3 people one question part A evaluate log 2 5 + log 2 1 over 40 so part A rules the logs again B same base with a plus means I can times these two together so that becomes log 2 5 * 1 over 40 so that is log base 2 of 54ths which is log base 2 of an eth cuz you can divide by five imagine as 2^ of something is an e well 2^ 3 is 8 so 2^ minus 3 is an e cuz it's 1 over 2 1/ 2^ 3 so that means that 2^ minus 3 = an 8 so this means that the answer is minus 3 Part B A is a member of the real numbers and that log 8 a is negative stay the possible values of a so we're saying that log base a of a is negative so less than zero we don't need to write v as so log base a of a is less than Zer let's just think about it logical for a minute let's imagine that we did 8 to the^ 0 we would get one if we take a negative number B we get 8 to the^ let's say -2 that's the same as 1 a^2 which is a fraction and if I take any negative number 8 to^ of a massive negative number would be a fraction so it's always going to be either as big as one or as small as almost zero so that means that zero is less than a is less than one we can see this on a graph I suppose if I draw a basic log graph there is one there and below the x axis is when it's negative so less than one is negative but only way up to zero it never touches zero so it's between 0 and one so that's another way you could afford it evaluating the expressions for logs s high mass 2016 paper 1 question 14 part A evaluate Log 5 25 to because remember that means 5^2 = 25 it's as simple as that part B hence solve log 4x + log 4x - x = log 525 so looking at our left hand side we combine combine that into one log by saying that it's x * x - 6 so I'll do that to start with log 4 x X - 6 and the right hand side log 525 well we know that's two so we can just write two so then we can use our rules of logs to say that it's a base to the^ of two equals the left hand side so in other words 4^2 = x * x - 6 they eliminate the log now we're just going to get a quadratic 16 = x^2 - 6 x moveing over to the same side you get x^2 - 6 x - 16 = 0 so you've got a quadratic to solve hopefully that's factorizable x and x 8 and 2 - 8 + 2 because - 8 + 2 is - 6 - 8 * 2 is -6 so that gives me x = -2 or x = 8 but x = -2 is not a valid answer because the log of zero or anything negative is undefined so X has to be greater than six so we can just score that out and say that X has to be greater than six is the answer so our final answer is X = 88 and we're done there solving logarithmic equations sqa higher math 2017 paper 1 question 12 given log a 36 minus log A4 = a half find the value of a so we use a l of logs to combine the logs so that can give me log base a and we've got 36 / 4 CU of a minus = 12 so log base e 36 / 4 is 9 equal a half so elimin the log a to the^ of half = 9 a to the half = 9 so that means it's square root of a we could write that as suppose equals 9 so if I Square both sides a = 81 and we're done there solving logarith equations s High m for 18 paper 1 question 11 diagram shows log 3x on a diagram in the answer book sketch a curve with equation 1us log 3x and then find the point of intersection between the curves which will be soled with logarithmic equations so 1us log 3x the minus part here tells me to reflect in the x axis and then the one part which I'll do in another color here tells me to then move up by one so first of all reflecting on the x axis this point will stay where it is but this point will jump down here and that would be at the moment as a as a point 3 minus one but now I have to move up by one so this point is going to jump up by one so that becomes 1 one and you can see this point goes up to 3 0 so I can just write the number three here getting rid of my intermediate working which you could use a rubber if you were doing this by hand and since it's reflected it's going to go down and through here so we can just draw that in down here through the Y and there we go it's a reflection in the X AIS okay Part B determine the exact value of the point of intersection simultaneous equations so for Part B we've got y = log 3x but we've also got y = 1 - log 3x so we can say that they are equal to each other and find X so log 3x = 1 - log 3x so taking the log 3x to other side you'll have log 3x plus log 3x so two of them 2 log 3x = 1 I can take the the two up as a power so I get log 3x^2 = 1 now I've got it in the form log of something equals a number so 3^ 1 = x^2 3^ 1 = X2 and therefore I can find x x = 3 to ^ half or 4 x = < TK 3 and we're done there notice it only be a positive part of this instead of a negative part because if x was negative then you would have log of a negative number which doesn't exist so it's just a positive I can take a not of that just positive value as X has to be greater than zero okay evaluate numerical expressions X high mass 2019 people one question 14 part A Val log 10 4 + 2 log 10 5 so we've got log 10 4 + log 10 5^ 2 I can take the two up so that's log 10 4 + log 1025 and then we can combine that because we can times the numbers together so it's log 10 4 * 25 which is log 10 10 100 log 10 100 is = to 2 because 10^ 2 = 100 and there we are Part B solve log 2 7X - 2 - log 2 3 = 5 so we'll combine that into one log straight away by writing log 2 7 x - 2 all over 3 cuz a minus means we can divide that equals 5 so then using our rules of logs to eliminate we've got a log on one side the number so 2 to^ 5 equals this so I can write that 2^ 5 is 7 x - 2 over 3 so we need to work out 2 to the^ of 5 or 2 2 is 4 * 2 is 8 * 2 is 16 * 2 is 32 so we get 32 so 32 = 7 x - 2 3 so we can times 32 by 3 32 * 3 is 96 so 96 is 7 x - 2 add two to both sides so 98 is = to 7 x dividing 3 by 7 then you get 14 = X or nicer x = 14 and we're done there sqe high math t22 paper 1 question 8 solve log base 6 x + log base 6 x + 5 = 2 where X is greater than zero so a couple of different methods you could have used here so I'll go through the alternatives for you let's call this one method one and the first method is to combine the two logs the plus is one single log so we could write that as log base 6 of x * x + 5 = 2 and there's your first Mark there okay moving to the second Mark we can write that as 6^ SAR = x * x + 5 or x * x + 5 = 6 so there's our second Mark there and then we can move on to try and solve that as a quadratic equation so standard quadratic way multiply the brackets we get x^2 + 5x and that equals 36 so tiing that upus 36 = 0 there's our third Mark there for putting it in the proper quadratic form and then our fourth Mark try to solve that double brackets x and x 9 and four times together to make 36 but take away make five so we've got + 9 - 4 so x = 4 or x = - 9 but X has to be greater than zero cuz it says so in the question so we disregard one of the answers so since X is greater than z x = 4 is the only valid answer and there is our final Mark right there we have to have stated that x = 4 we disregarded the minus 9 to get that final Mark let's look at method two so method two starts the same way we apply the same rule as before so we write log base 6 of x x + 5 = 2 and we get our first Mark the log base 6 of x x + 5 equals the log base 6 of 6^ squar cuz remember log base 6 of 6 is 1 so that is two and that would give us our second Mark and then we can cancel the logs from both sides to just write x x + 5 = 6^ 2 or X2 + 5 x = 36 and it continues in the same way as the last method minus 36 is zero there's our other Mark and then for our final Mark solving that equation we get x and x 9 and 4 + 9 - 4 = 0 so x = 4 or x = - 9 but since X is greater than z x = four is the answer and that's our final Mark there for four marks ex higher math 2023 paper 1 question 3 solve log 5x plusus log 53 = 2 so rules of logs if it's a minus of the same base you can divide these together so in one step we get log base 5/ X over 3 = 2 and then you need to get rid of the log so remember it's base to the power of this equals this so in other words 5^2 = X over 3 or to write it nicely X over 3 = 25 5^ 2 x / 3 = 25 so that means that X must be 3 * 25 so x = 75 and we're done there 2024 paper 1 question 9 Express log A5 + log Aus 2 log a in the form log a k where K is a positive number or integer so use the logs log base a of 5 and 8 is plus so 5 * 8 and then I'm dividing by something well let's take that up to the top here so let's do this one that's log base a of 10 2 so divide by 10 s so that gives me log base a 58 is 40 so that's 400 10 10 is 100 that's log Bas a of four and we to leave it in that form so we're done there exponential equations high mass 2016 paper 2 question six otherwise known as experimental data with exponentials that is stting the goow of a bacteria strain the number of bacteria given is B = 200 e 107t State the number of bacteria present at the start of this study the start means T is zero you can either remember then that gives you e nothing which is one so you get 200 or you can write that out actually so at the start tal Z so B of T = 200 e over 0.107 * 0 that's 200 * e a z which is just 200 calculate the time taken for the bacteria we to double for Part B so if it doubles that means that b of T becomes 400 so I can then write 400 is our B = 200 e if is 0.107 T was our equation and we need to solve that so divide by 200 400 200 is 2 which is what you should get e 0.107 t now we can take the log of both sides sides the natural log so log 2 or log E2 equal log of e 0.107 or if you prefer log e taking it down as a power so you get log 2 = 0.107 T * log e of e which is just 1 so log 2 = 0.107 T / 3 by 0.107 t = log 2 / 0.107 get your calculator out and you will get about 6.48 so 6.48 hours because hours was our unit of time soling exponential equations s high mass 2018 paper 2 question 11 a supermarket has been investigating how long customers have to wake up the checkout during in half hour period the percentage of customers who wait less than T minutes can be modeled by this equation P = 100 * 1 - e T part A a 50% wait Less Than 3 minutes determine K so I've got two values I can just substitute in so for part A P = 100 1- e KT so 50 is our percentage equals 100 * 1 - e^ of 3 K because T is three multiplying the bracket out we get 100us 100 e 3 K so we can take the 100 over 50 - 100 is - 50 = - 100 e- 3 K dividing through by - 100 - 50 over - 100 = e 3 K well that's a half so that's 0.5 = e 3 K you could have actually divided by 100 at this point here to get the half whated it from there but you get the same answer take the log of both sides natural log of 0.5 = the natural log of e 3 K so the natural log of 0.5 = 3 K the natural log of e so that means that log of 0.5 = 3 K so K is the log of 0.5 over 3 and we can work that out is a value in our calculator you get minus 0.231 and we're done there Part B says calculate percentage of customers who wait 5 minutes or longer so T is just five p is 100 1 - e - 0.231 t CU K is- 0.231 so when T is 5 minutes P = 100 * 1 - eus 0.231 * 5 and I would just need to work that out if you do that in a calculator you'll get 68.5 so it's 68.5% wait for less than 5 minutes CU remember this is a this formula is for the number of percentage of customers that wait for less than a pacific time so that means that for those who wait 5 minutes or longer which is what the question is I just do 100us 68.5 which is 31.5% wait for 5 minutes or more and we're done there soling exponential equations X high mass 2019 paper two question n a spacecraft can be produced by a type of nuclear reactor electrical power produced by a generator is PT = 120 e to Theus 0.0079 don't be put off by this T it's just part of the symbol to tell you that P is a function of time H where PT is electrical power and wats after T years determine the electrical power keyword initially produced by the generator so if it's initially produced by the generator that means part A when t = 0 PT = 120 e to Theus 0.0079 time 0 well that gives you P of T = 120 e to the power of 0 which is 120 because e to the power of Zer is one so we're done there we could put the units in just to be clear whats Part B calculate how long it takes for the electrical power produced by the generator to reduce by 15% so let's reduce this by 15% but a little bit of maths if it's reduced by 15% is 85% less so that means that repeat power is going to be 0.85 * 120 and you just get a calculator out for that to get 102 so that's our new power so that means that 102 equals the initial power 120 e to the minus 0.79 t and we're going to solve for T so it's an exponential equation so I can divide through by 120 to get 102 over 120 = e to- 0.007 9 t that gives me 0.85 on this side e to the minus 0.0079 t and if we're working in percentages you'll realize that this number here will be basically what the questions askes anyway but I always start from here just to be in the safe site so then we can log both sides using the natural log the natural log of 0.85 or log e of 0.85 equals a natural log of e remain is 0.790 put his standard stuff that means that log 0.85 equals take the power down minus 0.0079 * the log of e a natural log of e is 1 so that means that log 0.85 is just equal to- 0.0079 T and therefore T is a natural log of 0.85 ided - 0.0 079 and we can just get our calculator Place log 85 over 0.79 minus in front of course and you'll get 20.6 and the units were years 10 the hip tform was an Athletics contest made to seven events athletes score points for each event in the 200 M event points are calculator using this formula where p is the number of points and T is the athletes time in seconds calculate how many points are awarded for a time of 24.5 seconds in the 200 M event so we just need to substitute 24.5 5 and for T so we've got P = 4.99 087 42.5 - 2455 to the power of 181 that's 49908 7 time 17.95 to the power of 1.81 929 0368 let's just make that 99.04 it doesn't quite specify whether the points of whole numbers or not so that seems reasonable given the accuracy of the rest of the question right in the long Part B and the long jump event points are calculated using this formula where p is the points D is the distance and and CM and K as a constant it says 850 points are awarded for a jump of 600 calculate K so we'll just substitute n and work from there so we've got p which is 850 = 0.1 18887 D is 600- 210 to the power of K so 850 = 0.1 18887 390 to the k so dividing through we get 850 over 0.1 18887 = 398 to the^ of K loging both sides I'll just use the natural log taking the K to the front then equals log of 850 over 08887 / by the log of 39 so it's time to get a calculator to what GES out 1. 40999 so K = 1.41 M23 paper 2 question 13 a patient is given a dose of medicine the concentration of medicine the patient's B is modeled by C of T 11 e to Theus 0.0053 t for T is the time of minutes after the dose C is the concentration and G per liters calculate the concentration 30 minutes after the dose was given okay so for part A all we're doing is saying that c = 11 e even minus 0.0053 and that's going to be times on the top by 30 11 times e to the power of use my brackets - 0.0053 * 30 close my brackets press equals you get an answer of 9.38 29 so 9.38 will be fine and migr per lers where or not we'll take a mark off or not for losing the units you need to get units if you can okay b c time taken for this do to become ineffective and it says it do becomes ineffective when the conc force is 0.66 so all we're doing is sub then 0.66 to here and solving the resultant equation so for Part B I'm going to write 0.66 = 11 e is 0.0053 T so I can divide through by 11 straight away if I do 0.66 ided 11 I get 0.06 = e minus 0.0053 t take the natural log of both sides you can either write it as log e or RN I'm just going to write RN 0.06 = RN e minus 0.0053 t so RN 0.06 equal - 0.0053 t taking that to the front RN e but RN e is one or log e e is one so that means the r 0.06 = - 0.0053 t and then dividing through by the negative there I get Ln 0.06 / by- 0.0053 so just calculator 0.06 divided by minus 0.00 53 and we get 53.83 [Music] times in minutes so 531 or any reasonable rounding minutes make sure you like your minutes no need to change that to hours and minutes minutes should be fine High math 2024 people2 question 11 the number of electric vehicles worldwide can be modeled by Nal 6.8 e KT where n is estima number of vehicles in millions watch out for that word t is a number of years since the end of 2020 and K is a constant use the model estimate number of vehicles worldwide at the end of 20 so at the end of 2020 t = 0 so you get n = 6.8 * e 0 which is 1 so that's 6.8 but it's not 6.8 6.8 million you'll probably lose a Mark if you miss that word because it's not 6.8 Part B at the end of 203 is estimated 125 million Vehicles World R DET the value key so we want n is 125 million 125 = 6 8 e to the power of and K T So T 30 20 10 years 10K so that's a logarithmic equation it's going to turn we're going to use logs it's an exponential so divide by 6.8 that equals E 10 K take the natural log of both sides Ln right 1025 over 6.8 = Ln E 10 K 10K can come down to the front to get Ln which is just one remember and that is ln1 125 over 6.8 and therefore K is very simply ln1 125 over 6.8 all divided by 10 which is a calculator job Ln 125 5 over 6.8 all divided by 10 0.291 one blah blah blah I think 291 is fine 0.291 and that's K have we answered the question yes so we're done there graph say 2015 paper 1 question 13 the function f ofx is 2x + 3 is defined on r set of real numbers the graph of equation y f of x passes through 1 B and cuts the y axis that Q is shown what is the value of B well it passes through 1 B so I can just sub one in so for part A we've got f ofx is = to 2x + [Music] 3 so at x = 1 we're going to say that f of x = 2^ 1 + 3 which is 2 + 3 which is five so just the answer for question B = 5 and we're done there so Part B says copy a diagram above inspect inverse function and then write down the coordinates of the images of p and Q what it means by the images of p and Q is what does p and Q become in the inverse function so this is an exponential so the inverse function is a log so we need to reflect that in the line Y = X so let's just go ahead and do that so I've copied the diagram below so we already know know that P is 1 B which we just worked out as 15 and then let's work out our Y intercept for this graph so that is when x = 0 so FX which was 2^ x + 3 that's 2^ 0 + 3 1 + 3 is 4 so we get 04 we need to reflecting the line Y = X so draw a y = x sign so 0 four becomes a long 4 Z and 1 five becomes 5 1 and we've got an exponential curve like so that's the inverse function of X so that's part one done and then for part two I've already done the work for it the image of p and Q we've got 40 and we've got 51 and we're done there part C says r311 rise on the graph of equation y f ofx find the coordinates of image of of r on the graph of equation Y = 4 - FX + 1 so wrri down f of x again f of x remember was 2^ x + 3 and we know that the r lies on this which is 311 so our new equation is y = 4 - FX + 1 so what does that mean well let's just take a few notes in here means we move left by one minus f ofx means we reflect in the x axis and you can think of this that's like plus 4 so move up by four if it helps you can rewrite this as minus FX + 1 and then add the four in the end and you might be to see that B that means we've got our R which was 311 so doing the things in the correct order we we apply our refle and then we move the left and then we move up so our refraction minus f of x + 1 3 11 becomes 3- 11 cuz it drops down underneath and then that goes to moving left by one so three becomes two so 2 - 11 and then finally we need to move up by four so we get two 2 - 7 as our final answer now if you're really struggling with that one way to do it is to go back to the original graph and actually draw it reflect it move it to the left move it up and not that point moving around okay logarithmic graphs a s high mass 2016 paper 1 question 10 diagram below shows the graph of a function log 4X base 4 the inverse function exists on a diagram on your answer book sketch a graph of inverse function okay I'll just draw you a quick sketch of this so you got your y AIS you've got your x axis and you should know that a log graphs inverse is an exponential reflect in the line Y X so these points you can just switch the order so 1 0 becomes 0 1 and 4 1 becomes 1 14 so I can knowe 01 on the Y AIS and it was 41 so a long one up four let's just draw it there and there's are two points and then all you need to do is draw your curve and you can note that that is f of inverse f ofx f of -1x and we're done there soling log equations s High m for 18 paper 1 question 11 diagram shows log 3x on the diagram in the answer book sketch a curve of equation 1 minus log 3x and then find the point of intersection between the curves which will be solved with logarithmic equations so 1us log 3x the minus part here tells me to reflect in the x axis and then the one part which I'll do in another color here tells me to then move up by one so first of all reflecting in the x axis this point will stay where it is but this point will jump down here and that way be at the moment as a as a point 3 minus one but now I have to move up by one so this point is going to jump up by one so that becomes 1 one and you can see this point goes up to 3 0 so I can just write the number three here getting rid of my intermediate working which you could use a rubber if you were doing this by hand and since it's reflected it's going to go down and through here so we can just draw that in down here through the Y and there we go it's a reflection in the xaxis okay Part B determine the exact value of the point of intersection simultaneous equations so for Part B we've got y = log 3x but we've also got y = 1 - log 3x so we can say that they are equal to each other and find X so log 3x = 1 - log 3x so taking the log 3x to the other side you'll have log 3x plus log 3x so two of them 2 log 3x = 1 I can take the P the two up as a power so I get log 3x^2 = 1 now I've got it in the form log of something equals a number so 3 ^ 1 = x^2 3 ^ 1 = X2 and therefore I can find x x = 3 to^ half or X = the sare otk of 3 and we're done there notice it only be the positive part of this instead of the negative part because if x was negative then you would have log of a negative number which doesn't exist so it's just a positive I can take a note of that just positive value as X has to be greater than zero higher Mass 2023 one question 9 the diagram shows the graph of function f of x is log base 3x X is greater Z of course rers function does exist on in the diagram in your answer book draw the graph of the inverse function minus one well let's examine inverse function first the inverse function of a log is an exponential and essentially the easiest way to draw it imagine a doed line here y = x the coordinates just switch places okay so here is my normal y = x so what 1 0 becomes 0 1 and 3 1 becomes 1 3 and it just giv you an exponential curve and that would be 13 so that's the inverse function of X so now I just need to take away one which means just to move it down by one so one is going to go to zero and 1 three is going to go to one two so I can just draw it now there's zero it's obviously going to drop below because that's where your ASM toop is essentially so it's going to go along up through zero and start climbing and we just put that point on that we noted which was 1 three so it now becomes still one but two and we're done there F finding from a straight line so now I was changing the axis so that one or both become logarithmic axes and then that makes a straight line so we can work from there here so let's have a look at this question two variables X and Y are connected by the equation y = k * x^ of n the graph of log 2 y by log 2x is shown find the values of K and N so I don't me you can memorize how to do this but I usually just start from the beginning which is to say that y = KX to the N if I've got y = KX to the N I can log both sides and the base is two so I can say that log base 2 of Y equal log base 2 of KX to n separating the product K and x to the N I can say that log 2 y equals I'll write the X One first log 2 x to the N plus log 2 K so I can then write log 2 y equals n comes down log 2x + log 2 K and now looks like now y = mx + C that's like your b y equals that's like your m x + C so our gradient is equal to our n and our c is equal to our log 2 K so if we can find the gradient we know our n and we can find C we quick cut to Y AIS we know log 2 K so let's look at our graph so for our grad we've got along 12 and up 3 3 over 12 so our gradient is 3 12th so our gradient is equal to 3 over 12 which simplifies to 1/4 and therefore n = a quart now to get our C we know that log 2 K is where it cuts a y AIS our graph tells us it cuts the y axis with number three so we're just solving log 2 K = 3 that means 2 to^ of 3 = k 2 to^ 3 = K and therefore K = 8 CU 2 cub is8 and we're done there it from a straight line for logarithmic graphs s high mass 2019 paper 2 question 12 two variabl X and Y connected by equation and this is what you're looking for y = AB VX a power and we're looking at one of the axes is logged but the other one is not so it's only one log this time but always start from your equation and get the logs coming out and see where you go so I'll start with y a b VX we to find the values of A and B so let's start with y = a v x just log both sides in our log Bas is four so looking on our graph log base 4 y = log base 4 of a v x which is our product so I can separate that into a plus so log base 4 y equal log base 4 of B VX I write the X1 first MX plus C plus log base 4 a taking the X to the front we get log base 4 y That's a y and that's a log = x * log base 4 B plus log base 4 A well let y = mx plus C is our straight line so let's put the X on the end so you can really see it log base 4 y = log base 4 B * X Plus log base 4 a so that's like y = mx + C so we can see our a gradient is equal to this number in front of X and our c is equal to S so we can say our gradient is equal to log 4 B so let's just check our graph to see if we can work out the gradient we're given two points so our gradient is equal to Y2 - y1 / X2 - X1 so that's 8 - -1 over 3 - 0 that's 9 over 3 which is 3 so we know our gradient three so log base 4 B = 3 remember that means that 4 to the^ 3 = B 4 cubed is B 4 cubed is 64 and therefore B = 64 now let's try our C form work we've just done our c is equal to log base 4 of a c is where it cuts the y axis we know it cuts the y- AIS at 0 - 1 so C is min -1 so we can just straight away save it -1 is equal to log base 4 of a or if you prefer log base 4 a = -1 remember that means 4^ -1 = a just write that straight away 4- 1 = a 4^ - 1 is 1 4^ 1 so therefore a = one4 and we're done there question s says two variables X and Y connected by equation y = KX n the graph of log 5 Y and log 5x is shown as a straight line calcul values of K and N and we're obviously given two points so let's start this question so we got y = KX to the n and we've got the log base 5 so we can log both sides with base 5 so log base 5 y = log base 5 K x n which we can separate into two separates log base 5 y y = log base 5 x n plus log base 5 K bring an N down in front we've got Log 5 y = n log 5x + log 5 K that's a bit like because it's logarithmic log base 5x and log base was y that's a bit like a big y equal m m x + C so that's our gradient and that is our Y intercept so we can work out our gradient because we're given two points in the question just remind ourselves 03 and 2us one so our gradient equals minus 1 take away 3 over two take away zero that's - 4/ 2 which is min -2 we can then do our Y intercept that's C remember so C was log 5K that cosses the Y AIS at the number three so that means that 5 Cub = k so25 = k so to answer our question we had y = KX to the^ of n so Y is 125 x the- 2 where K = 125 and N = -2 higher math 2024 two question 6 two variables X and Y are connected by the equation y = a x b the graph of log 5 y log 5x are shown and it's a straight line so we've got Log 5 y by log 5x and fin the me right so I always usually start with equation some people skip ahead but I don't like to so start with that and I log both sides the Bas is five log base 5 y = log base 5 ax B so now I can separate my log out on the right hand side to log base 5 a plus log base 5 x b so that means log Bas 5 y = log base 5 a plus b comes out the front log base 5x and putting it in a kind of form of y mx plus C you've got your log base 5 y I'll take the X bit first so B * log base 5x plus log base 5 a so now just comparing your graph your y AIS is log base 5 y so that's like your Y and your xaxis is log base 5x so that's like your X so that bit there's your gradient MX that bit there as just C so the the gradient is B so what's the gradient of this line we I've got two points 0 - 2 and 4 10 so the gradient between 0 - 2 and 410 is what we're going to do so my gradient = 10 - -2 4 - 0 that's 12 over 4 that's three so that means it to answer the question B = 3 now it says find C well I already know where it cut to Y AIS 0 - 2 so that's it C is-2 so if C is min-2 that means that log a log 5 a = -2 so that means that 5 to^ -2 = a that's 1 over 5^ 2 5 S is 25 so a is 125th and we've got that we've got that so we're done this been quite math today we' done through the whole of logs in higher maths all the logarithmic functions all exponential functions log equations exponential equations graphs of relating functions experimental data take care stay safe and goodbye