okay you have made it to the end of pure year 1 this is chapter 14 and we're looking at a really big topic which is exponentials and logarithms and it comes up a lot in year 2 as well so make sure you really know what's going on here so I'm going to begin by having a look at stuff to do with the graphs and some stuff to do with exponentials and solving equations now really you just need to be aware that you do need to do graph Transformations um for this particular topic and I'm going to just kind of name what some of these graphs are and try and talk a bit about like the idea of exponential growth exponential to K and things like that so if we start off with sort of the most traditional looking of these this is what an exponential graph looks like these are all just going to be examples so I'm just going to say this is an example of something like Y = 2 to ^ of X we've got an exponential growth now for this one that we have down here we've got this green one this green one could be written in a couple of different ways actually we could write this as something like I don't know y equals a half to the power of X because if the base is less than one or between zero and one we get this exponential decay or I suppose it could be written as Y = 2 ^ of - x and this should make sense because the 2 to the power of a minus is the same thing as a reciprocal of two which is a half so I'm kind of starting to see how this connects to some stuff to do with graph Transformations as well and then these ones that we've got down here I suppose I'll probably start off with this blue one this blue one that we've got it's like the same as this red one that we've got here but it has been negated on the outside so it's like a y equal minus 2 to the^ power of X like this and then for this purple one it's like one of these ones that has just been negated as well so I could say it's either one of those two negated I'll probably just go with this one y is minus a half to the power of X but equally you could have written it as - 2 to the^ of - x now this graph that we've got here this is the graph of y = lnx okay this is yal Ln X or I suppose it could also be y = log X just some kind of logarithm and what you'll see in pure mat year 2 is that this is the inverse function of an exponential function and it's actually just an exponential function that has been reflected over the line yal X so I mention these here just because I want you to be as aware as possible that you do need to know the shapes of these graphs in general now this gets kind of in a little bit more depth well quite a lot more depth about differentiation in year 2 but for now you just need to know these facts that if you have the function yal e to X then to differentiate it the very special thing about e to the X is it just differentiates to itself it differentiates to e to the X this is going to be a particular thing I'm just going to tell you but in year two you'll see that this is an application of the chain rule so if we had yal e to the^ of KX where K is a constant it just differentiates to k e to the power of KX so whatever the coefficient of x is it comes out to the front and in year two if we didn't have it as e as the base so we had any number like 2 the^ of X then to differentiate it you get Ln of that value multiplied by the same thing so e to X goes to e to X because it would be Ln e * e to^ of X and hopefully you know that Ln is equal to one now it's probably worth me mentioning here these videos aren't going to teach you everything they're going to be here as a refresher so there may be a little bit about stuff with logarithms that aren't being mentioned here for example that Ln is equal to one but hopefully you've covered that through watching my videos too now some of the bits we need to remember for the log laws if we add two logarithms that of course must have the same base here then we can actually multiply their inputs so that we get the log of a and you can put brackets around it if you want to if they are being subtracted it would be the log of a divided B and this one we know that if we have the power that is at the top you can put the power to the front so that we get B log a and of course all of these laws can be used in Reverse so you can split two things and make an addition you can take a division and split it into a subtraction and of course you can take this thing here and put it inside as a power now remember that you cannot input a negative into a logarithm so you need to check your answers to see if they are valid and that's because if you look at this graph of a logarithm there are no values there's no um graph for any of these negative values of X it only exists for values of X greater than zero you can't even input zero into a logarithm um so let's actually try this last part which says we also want you to look out for pseudo quadratics sometimes called quadratics in Disguise when you're looking at exponentials so really this is probably one I wouldn't want you to just watch this video to teach you it in fact I wouldn't want you to do that for any of them but this playlist I've made on exponentials and logarithms will really teach you why they work and if any of you are watching this from year 13 you will know how much exponentials and logarithms come up in year 13 maths as well so we're going to begin doing a pretty easy one we are just going to start off by solving this equation which is 4 ^ of x = 7 now there's two ways of doing this you could just go straight in with its definition so we know that we're saying that X is the power of so log means the power of the power of Base 4 that gives me the answer seven log base 4 of seven I can do that on my calculator if we wanted to get it as a numerical value and this is 1.44 and that's to three decimal places now there is another way you can do this that you might see me using a lot more in my videos you can just instead of taking log base 4 you can take Natural logarithms of both sides so you get Ln of 4 ^ x = 7 well that X can now be written outside the front so that we get X ln4 oh I didn't take the logarithm of both sides that should have said ln7 so if I just finish this off I get that X is equal to ln7 / ln4 and just on my calculator I will do ln7 divided by ln4 to confirm it gives you the same thing as 1.44 to three decimal places so this this form that we've got here and here are equivalent one is in terms of log base 4 the other one is in terms of natural logarithms should you prefer to do either it's absolutely fine and you should probably know both approaches now this one we've got 2^ x equal 3 power of 2x + 1 and we want to give our answer in a natural logarithm form which means we must take Natural logarithms of both sides for this so if I take natural logs of both sides I get Ln of 2 ^ of x equal Ln of 3^ of 2x + one and we can now use the log l of pulling these Powers down and writing them at the front so that we get X Ln 2 equals must have brackets for this one 2X + 1 Ln 3 and remember I'm trying to find out what x is equal to so I'm going to need to expand these brackets so I can eventually gather all of the X terms together so X ln2 is equal to I will get a 2X Ln 3 plus Ln 3 now I think I can subtract this to that side so I gather all of the X's together so that is x ln2 - 2x Ln 3 equal Ln 3 I've got two x's so I'm going to get them as factorizing just the One X out so I'm left with 1 x so that's x ln2 - 2 Ln 3 = ln3 my final thing is just to divide so that I get ln3 divided ln2 - 2 Ln 3 which I hope was the form they wanted yeah so we've got the a is 1 B is 1 and C is minus 2 for this one that we have Okay so so this time we've got solve 5^ x + 1 = 5 ^ 2x - 6 giving your answers to three decimal places now I'm going to do something that is totally wrong you might think this is a good idea let's do this let's do this let's do this now we can't take logarithms like this if you take a logarithm just like squaring you would have to take a logarithm of the entire thing so that it would be written like this and then we get a bit stuck there because we actually can't split this up so this is not one where we're going to take logarithms it doesn't work and I mentioned it somewhere over here I said look out for pseudo quadratics with exponentials so I think in this particular case what we're going to do is a little bit of manipulation let's see if we can think about what this might be the same as it is a 5 ^ of x + 1 that's a 5 to ^ of X multiplied by a 5 to the power of 1 we then have a 5 the^ of X all 2ar - 6 X so if I was going to and you don't have to do this you could go straight in if I was going to let y = 5 ^ x I have y * 5 = y^2 - 6 so we have a quadratic it's a pseudo quadratic so that's a y^2 - 5 y - 6 I can put this on my calculator or I can just solve it this is a y - 6 and a y + 1 so either Y is equal to 6 or Y is equal to minus1 now remember we do need to finish this off so we're now saying if Y is equal to 6 we have that 5 ^ of X is 6 and so X is the logarithm of Base 5 that gives me six I'll put that on my calculator log base 5 of six you could do the Ln method if you wanted to and it wants me to give our answer to three decimal places that's 1.13 and then this one if I just pull it aside so I have some more space we're actually saying that 5 ^ of X is equal to minus one now if you just think about what we said to do with graphs this is what an exponential graph looks like can this exponential graph ever be equal to anything down here a negative no it cannot so it has no Solutions which means this equation only has this one solution of 1.13 because there is of course this ASM toote that we have here this is I should have perhaps said this is an ASM toote here and this is also an ASM toote here for the logarithm graph that we've got then a nice easy one that's not got much space at all because I knew we could squeeze it in nice and easy we are just going to differentiate this so we get that Dy by DX is equal to the K in this case is min-2 so it would be the 4 * by the minus 2 which is - 8 and then it just stays the same e to^ of- 2x that's just an application of this part that we've talked about here so let's do some solving equations with logarithms we're going to be using some log laws with this and I'm going to begin by looking at this one now you might think you immediately want to combine them that you want to do an x / 3 but because there's a two here and not a two here of course we cannot do that if they both added two you could combine them but we can't so the first thing is I'll put this two inside the logarithm of base two so it'll be an X2 then I will combine them because there is a subtract so I can do it as a divide so that's going to be an x^2 / 3 and now I'm going to rewrite this statement because it's now as a power statement I want to write sorry it's a logarithm statement I want to write it as a power statement so it's telling me the power of base 2 that gives me x^2 over 3 is 5 so it's telling me the power of base 2 is five to give me X2 over three the alternative to doing this if you've watched my videos you'll know this is you can say okay well the opposite of log base 2 is to do two to the power of so I can say that this is equal to this the opposite is doing 2 to the power of on both sides when you do two to the power of it gets rid of that part okay I'm going to erase this CU I don't want this to stay on the annotated version so I now have got that x^2 over 3 is 2^ 5 2^ of 5 is 32 so I will do my calculator 3 * 2 ^ 5 that's 3 * 32 which is 96 so 96 is x^2 so we take the < TK of 96 which is plus or minus 4 < TK 6 but are they both valid well remember you cannot put negatives into a logarithm we said that here you cannot input a negative into a logarithm so although it looks like there are two answers we are going to reject uh - 4 < tk6 so X is equal to 4un 6 only it needs to be clear if you reject one of the answers okay we'll try this second logarithm one that we've got here now these two actually do have a two in front of them okay and there's a maybe an alternative way we can even do this I might see if I can talk to you about that in just a second so I am going to begin by putting these ones on the same side as each other so that I have ln4 is equal to 2 Ln x + 1 - 2 Ln of x - 2 so I'm going to just work on the right hand side so this is an Ln of x + 1^ 2us an Ln of x - 2^ 2 so we get that ln4 is Ln of x + 1^ 2 over x - 2 2 we've got some subtracting and we know when you do the subtracting we can write it as a division now because they've both got logarithms it should naturally make sense that four must be the same as this thing we can just remove those logarithms so 4 is equal to x + 1^ 2 over x - 2^ 2 I'm going to multiply by that x - 2^ 2 and I'm also going to expand those brackets at the same time so that's an x^2 - 4x + 4 and on the right hand side I'll expand this x + 1^ 2 which is x^2 + 2x + 1 so we have a 4x^2 - 16 x + 16 is x^2 + 2x + 1 now we're just back to doing quadratics aren't we so I subtract the X2 I will subtract a 2X and I will subtract a one and then this one I'm definitely just going to do on my calculator so going to my polinomial solver I'm putting the 3 the minus 18 and the 15 we get that either X is equal to 5 or X is equal to 1 now let's look and see if both of these are valid if x was 5 5 - 2 that's positive 5 + 1 that's positive but if x was 1 we would have 1 - 2 which is negative so we cannot have that one it would not work so I'm going to reject this one that we have here my only answer is x - 5 now I did say I can show you an alternative way of doing this because ln4 we can write in a different way ln4 is Ln of 2^ 2 which is pull the power down to the front 2 Ln 2 so we're left with 2 Ln of x - 2 + 2 Ln 2 = 2 Ln of x + 1 and now hopefully you've noticed they all have a two and so I also I should have said here I didn't have to put these on the other side I could have put the four with this or I couldn't because of the two so no ignore what I'm saying there ignore that now I've been able to cancel that two out I can solve this so I can actually combine these two as sort of multiplying together so that I have Ln of 2 x - 2 remember if you add them you can combine them like this and now that we have them both with these LNS I can remove the Ln from both of them so I'm just left with 2 x - 2 = x + 1 don't need the bracket anymore so that's 2x - 4 = x + 1 I'll then subtract the X from both sides and I will add on the four and we get the only answer of x equals 5 it doesn't give us this answer that doesn't work here so I thought I'd show you both of these because I really want you to get a deeper understanding of exponentials and logarithms this was a smart way of recognizing that ln4 can be written in such a way that we can then cancel through with the two okay and then we've got one that's sort of with some other variables we've got one with some exponentials here so we have that a is equal to 3 e to the 2 T minus 1 find the value of T when a is 13.5 so we get 13.5 is 3 e to the 2 T minus 1 pretty simple here we're just going to do our 13.5 divid 3 so there's my 4.5 is e to the 2T minus one and we just need to know that the inverse of doing an e to the power of something is Ln so I will take Ln of both sides now if you wanted to you could do Ln of e to 2 tus 2 you could pull that 2T minus 2 here pull it down to the front and actually Ln e is one so it just goes or you can just recognize that taking the natural logarithm the Ln of both sides here it literally cancels out that e so we just get left with the 2T minus one I realize I said minus 2 but it was supposed to be minus one so what I need to do is add the one on here and then divide by two and that will tell me what my value of T is so I'm going to do the Ln of 4.5 plus the 1 divide by two and I didn't say how many decimal places but I'm just going to do it to three significant figures so it's 1.25 to three significant figures and of course you can always take this 1.25 type it back in here and it should give you something close to 13.5 so I'm going to do two more questions one to do with modeling with exponentials and then one for logarithmic and nonlinear data on graphs now in this particular question they have given us the model but if an exponential model isn't provided the ones you should use is AB to the^ of X or just AE to the oh this is a mistake here this should say a e to the KX I think it should say um whatever it is going to say in the the final document when you use it I think it should be AE to the KX and I think this one is absolutely fine this one we can leave like this but this should be AE to the KX it will be correctly written in the PDF when you download to use it which is always in the description so we have the popul population P measured in thousands of a town T years after January the 1st 2010 is modeled by this equation using the model find the population of the town on January the 1st 2015 okay so really in these modeling questions all you can really do is either substitute or solve or comment on something so if it's the 1st of January 2010 and we're doing the 1st of January 2015 that sounds like T is equal to 5 so I can just do in my calculator 52 - 15 e to the - 0.15 multiplied 5 so because it's 5 years so 52 - 15 e to- 0.15 * 5 and this is 44914 now remember that's in th000 so I'm going to multiply that by a th000 and I'll say so the population is 44,990 and that is to three significant figures they usually say on the front of the exam paper that three significant figures is fine now for Part B it says find during which month and year the population is expected to exceeded 48,000 so 48,000 we know is that P is equal to 48 well month and year it's asking what the time is so we're literally just going to substitute that P is 48 and solve an equation so if p is 48 we get that 48 is 52 - 15 e to the- 0.15 T let's solve this I'll put this onto the right hand side so that's 15 e to Theus 0.15 T and then I'll do my 52 - 48 Now 52 - 48 is 4 and I'll divide by the 15 so I get 4 over 15 e to- 0.15 T now I'm going to get rid of this e part by doing Ln so that I get minus 0.15 t is the Ln of 4 over 15 and so T is the Ln of 41 15 / - 0.15 pardon me so the Ln of 415 divid byus 0.15 and we get that t is 88117 years now unfortunately that doesn't answer the question because that is in the years after January the 1st 2010 but we want to say which month and year it is expected to go to that population so let's just deal with the fact that this is is 8 years and then we're going to do 0. 8117 * by 12 to find out how many months it is so I'll subtract the 8 times that by 12 and we get that it's 9.74 months now be really careful here because if it said that it was 8 Z let's say sorry if it said 8.00 when we did this bit here that would literally be referring to jary the 1st 2018 and if it said 1 month afterwards it would be referring to February so we have to be really really careful with this we want to think what is 9.74 months after January the 1st well nine months after it would be I'm literally going to count through it' be February the 1st March April May June July August September October so 9 mon months after January the 1st is October the 1st so 9.74 months will be taking place during October so we can say that it will be October 2018 it looks like September because of the 9 but it's actually the 1st of October and then this 74 of the month means that we're going even further into October 2018 so October 2018 is when it will exceed 48,000 it then says for part C show that the rate at which the population was increasing on January the 1st 2013 was approximately 1,430 people per year so we are now talking about rate rate means differentiate so the population is 52 - 15 e -0.1 5T let's differentiate this is going to differentiate to DP DT the 52 is going to go to nothing we're going to have the minus5 and it's going to get multiplied by the K part up here which is minus 0.15 and the E part all stays the same so on my calculator I'll do my -5 Times by my -0.15 which is 2.25 e to Theus 0.15 T now they're talking about in January the 1st 20 13 so T is equal to 3 and I'm just going to type into my calculator 2.25 e to -0.15 * by 3 so 2.25 eus 0.15 * 3 and we get 1. 1434 like this but remember this is in thousands so I'm going to say 1.43 4 * 1,000 is equal to 1,400 34 which is approximately 1,430 people per year so the rate we were talking about the key word there was rate it then says state with a reason the limit of the town's population according to the model now the limit often makes us think what happens to this model as T becomes a very big thing you can either just type into your calculator that T is like 100 million years or a thousand years well you can start to think like this okay so I'm going to look at this model that we've got here I'm going to say as T goes to very big values e to Theus 0.15 T it's exponential decay it's going to go to zero hence this means that the population is going to go to 52 minus 15 * 0 52 - 15 * 0 which is just 52 so the limit is 52,000 people and we've actually given our reason here with this mathematical notation that we have okay the last kind of question you need to be a aware of is logarithmic and non nonlinear data on graphs there's often a couple of these questions by the way there's often like at least two or three across the papers it's a really really big topic so I'm just going to talk you through the steps for this but really just watching it is going to be the best way of seeing it so logarithms can be used to change from exponential or nonlinear data to linear data we can then use the math of straight line graphs to solve problems so the first thing we're going to do is take logs the second thing if we need to is find the equation of the line the linear line if necessary compare both equations solve for the unknowns and then there might be some extra parts to the question as well so this question says the value V pounds of a printer T years after it was bought is modeled by V equal AB to the power of T so we've got an exponential model here you can have different ones but taking logarithms is always going to be the trick and of course A and B are constants here when log 10 V is plotted against T like this like a graph here's our log base 10 of v and here is our T the relationship is linear and the line passes through the point 0 2.5 and 72 so actually it's crossing here at 2.5 or 0 2.5 and then it's also Crossing at another point which is 72 and it tells us that it's linear here using these points find a complete equation for v in terms of t giving A and B to three significant figures now the first thing I suggested we do is take logs of this model that we've got so my model is V equal a b to the T well when I take logs I can never be bothered to write the base 10 so I just write log log V is equal to the log of ab to the^ of T now you know your log laws so we can split this into a log a plus a log of B e to the power of T I'm going to save some time and I'm just going to take that power and put it to the front like this now you can also see from this graph here we've got a straight line graph it would normally be Y and X isn't it so I think what I need to do is find the gradient of this line find the Y intercept and then I can put it back together in an equation so the gradient is looking at these coordinates it's the change in y so I can do my I always do the second one minus the first one so I'm going to do the 2 minus the 2.5 and on the botom I would do the 7 minus the 0 so that's going to be a - 0.5 / 7 or -14 which means the line is normally going to be y = mx + C but we're not doing yal mx + C the y- axis is log base 10 of v and the x- axis is T so I've now done Point number one which is to take logs Point number two which is to find the equation of the if necessary point three is to compare both equations we do a bit of Spot the Difference well look they've both got a log V fantastic and then they've both got a t the thing that is next to the t is the log B so immediately I can see that log B is equal to -1 over4 which means that the remaining part of the equation the log a must be oh the plus C I didn't even say what the plus C was we can tell from the graph what the plus C was can't we sorry guys the plus C is the 2.5 it's a 2.5 so I can tell very quickly now that my log a which is in purple is 2.5 now you might want to remind yourself these are base 10 so B is 10 the power of -1 over4 and a is 10 the power of 2.5 so they want these to three significant figures yeah three significant figures so on my calculator I will do 10 to the power of -1 over4 which is 0.848 and a is 10 the power of 2.5 which is 316 so the equation is V = AB to the T A Times by B to the T remember this is talking about the cost of a printer so when the printer was bought it cost £316 when T was zero and we can see this 0.848 means that every year it is going down it's about 85% it means it's decreasing by 15% 85% means decreasing by 15% so for Part B of the question we are going to answer this thing it says exactly 5 years after the printer was bought AKA T is equal to 5 the value of the printer was 00 use this information to evaluate the reliability of the model so I'm going to use our model which is that V is equal to 316 * by 0.848 don't worry about those rounded values they've already told us to round it earlier on the question so it's perfectly fine to do I will do 316 * 0.8 for8 to the^ of 5 and we get that our model predicts that the value of it is £ 13857 so this is from our model now the real cost was £00 so I can say this is very far from the true cost and in brackets I can say it is a nearly it is roughly 39% overestimate obviously 38.572786 how close does it have to be this last little thing to finish off the video If a model is within 10% of The observed value we can say that it is a good or um reliable model otherwise we say that it is not good or unreliable so in this case if the um the model predicted something between like 90 and £10 we could say that's a good model because that's 10% either side of1 but in this case this was too far away so not a good model now this topic is huge okay these questions I've got here are not going to be enough to cover all of the different different things that can be asked if I did every kind of thing that could be asked this video would be twice if not three times as long so go and do lots and lots of practice on this you have now hopefully finished all of pure year 1 and I hope you've been finding these videos useful if you do find them useful and you're not subscribed subscribe to the channel like the video all of these things are helping me reach my goal of you need being able to teach as many students as possible but also I want to reach 100K subscribers so wishing you the best of luck with your studies and I hope to see you in another video soon