Transcript for:
Exponential and Logarithmic Functions Overview

hey what's up guys in this video what i want to do is kind of do a condensed version of everything you kind of need to know for exponential and logarithmic functions test so a lot of times when you know the day before my students would take a test we would do something like this where i just kind of do like a quick little review highlight some you know important concepts that i want my students to understand as well as kind of highlight some mistakes or some things to look out for so we're just going to kind of run through all the topics that i would expect for you to know on a test for exponential and logarithmic functions we'll do some specific examples and go through some specific problems we'll do some specific examples but also kind of talk broadly about a lot of the concepts to make sure you just kind of have a general idea so if you have a test coming up or you just need a refresher on exponential and logarithmic functions then this video was made for you now the first thing i think that you know we need to understand is what exactly is an exponential function so in that case what exactly is an exponential function so when we have a function like f of x equals b to the x okay now this is called exponential function because again what we notice here is we have a base right that's why we use the b and then we have the x which is going to be the exponent now it's important when we're thinking about an exponential exponent exponential function there's really kind of two different graphs we're going to be looking at the first one is going to be a growth and then the other one is going to be a decay and it's really kind of important just to make sure we recognize what exactly these graphs look like so therefore we can understand their behavior so the growth function is basically going to be the graph as it's going from left to right is going to be increasing now there is a horizontal asymptote which we'll talk about here in just a second and this is going to happen whenever my b is going to be greater than 1. now my decay function is going to look very similar to the growth function however when we're going from left to right we're actually going to be declining here and again we're going to be approaching this horizontal asymptote now there's one distinct point that we need to make sure we know and again this is very important when we're dealing with our graphing is going to be the y-intercept so the y-intercept for both these functions is going to be zero comma one and when we want to eventually graph our functions what we're simply going to do is just kind of shift our y intercept using our transformations or we'll shift using the asymptotes now the k functions are going to happen when i have a b which is going to be greater than zero but it's going to be less than one so typically that's going to be like a fraction but again don't get caught up with a mistake just because you see a fraction then it has to be less than one because there are fractions that are greater than one like five thirds or something like that remember if you can take the fraction and write it as a mixed number it's going to be larger than one so just a quick little example here you know if i had like y equals a three to the three fourths raised to the x a lot of times what we'll say is you know is this a decay function or is this a growth function so remember this is a exponential equation right exponential we are looking at the base this 3 right here is just going to be a scale factor that's actually going to be stretching the graph so we don't care about this the only thing that we care about is the 3 4. since 3 4 is less than 1 but it is greater than 0 then this is an example of a decay function so if i had these swapped right for instance if i had y equals a 3 4 you know times a let's say 3 to the x and wrote it like that it doesn't really matter but you can see now that my base of my exponent of my exponential equation is three and three is going to be larger than one so therefore that would be an example of growth so it's just very important when you're trying to identify if it's growth or decay to make sure you're looking at the b don't get distracted by that coefficient now if we have a function right our parent graph is y equals b to x we need to know exactly what that graph looks like right we need to understand our transformations and again just to go ahead and graph this and we'll assume that b is going to be larger than 1 here so therefore we can go ahead and draw this graph is going to be just like this okay and again remember we have this point here 0 comma 1. did i actually do that correctly yes i did okay now the next thing we want to do is what about the transformations what is happening to the graph when we multiply by certain numbers like when we add or subtract it like for instance what is this three-fourths due to this graph what does the three do to that graph right and what about if we add and subtract numbers so in our study of like transformations in other chapters we should hopefully recognize that whatever we transform from one function is going to be the same for even our exponential and logarithmic functions but we'll get the logs in just a second so let's just focus on the exponential here so i'm just going to use again some more letters if i had f x and we'll put an a in front here b and then we'll just do an x minus c plus d okay so remember a is going to be your stretch and compression so remember when a is going to be less than one that is going to be a reflection about the x-axis remember c is going to be your shift left and right and d is going to be your shift up and down now just remember whatever the value is of d like if d is positive if you go up if d is negative you go down and just remember c it's actually the opposite of c inside the function so if i have like x minus c that's going to be actually shifting it to the right and if it was x plus c that actually be shifting it to the left and again we can practice a little bit more of this like by using desmos and kind of seeing how that relates but exactly in this example like since we're trying to you know just review for the test i don't want to go into the table because when you're taking a test using a table is not always going to be the most efficient way to graph a function but if you get stuck then i would definitely recommend you know still kind of going back to those table stuff that we talked about in earlier lessons but really it's really important to recognize these and and also the importance of b now b is going to impact the graph as far as like how fast and slow it's going to be increasing but the main thing is it doesn't really matter what b is but the main thing like if you're just trying to get a general idea of what the graph looks like it doesn't really matter what the if b is like if b is two if two b is three like the general shape of the graph is going to be the same as long as b is greater than one obviously when b is less than one it's going to take a different curve right it's now going to be a decay function so b is not going to be a general transformation it is going to impact the graph but it's actually not going to be transforming or shifting you know the graph in that sense so let's go and take a look at at least some of the graphs let's go and see if we can transform them let's see if we can graph them let's talk about let's go back and review the domain and range the end behavior as well as how to identify some points on the graph okay so in this first example um the main thing we want to be able to do is just kind of understand our transformation so whenever i see an exponential function and you know if i'm trying to graph it or if i'm trying to identify like the domain and the range i still want to be able to identify the transformations so the first thing i'm going to look at is i'm since i'm subtracting one inside the function right it's inside the function because it's listed as an exponent that's going to be a shift one unit to the right and since i'm adding a 6 on the outside of the function right it's not actually in the exponent that's going to be shifted up 6. so my transformations are going to be shift right 1 and shift up 6. okay now so if we want to sketch this graph basically all we need to do is go ahead and take the paragraph and then just apply some transformations right so my parent graph here looks something like this now so if i want to find the new transformations again what am i going to do i'm just going to shift the graph over 1 unit to the right and then up 6 units so i can take this point shifted up 1 and then just go up 6 units so 1 2 3 4 5 6. so my new y intercept is going to be right here now again you have to remember there's a horizontal asymptote going here so in addition to the x-intercept being shifted up six units also my horizontal asymptote needs to be shifted up six units and again this is not visible but it's important because that's what the graph is approaching so we're going to go up one two three four five six and you can say here i'm going to have a horizontal asymptote right here now you can see this graph is going to take this kind of shape right so let's go ahead and move that down because it's kind of like in the way but hopefully you kind of see what's exactly going on here and actually i'll just fix this so then you can see the graph is going to be taking a shape of something like this which is basically the exact same shape of my previous graph it just got shifted again one year to the right and then up you know six units now typically i do like to be able to say like what is this value like how do you find the y-intercept like whenever you're graphing it's usually is helpful to have at least two points and um the easiest points to find is in addition to the old y-intercept right just by applying your transformations we can also go ahead and find the y-intercept and again the y-intercept is going to be when our x is equal to 0. so to do that what if i want to find the y-intercept i'm just going to say then x is going to equal to 0. so if i have f of 0 is equal to a 3 times 0 minus 1 and then that's what plus 6. all right so let's see what this is going to be equal so therefore i have 3 to the negative 1 so it's going to be 3 to the negative 1 plus 6 so that's going to be a 1 3 plus 6 and then again if i want to get these to be common denominator just remember that's going to be over 1 so then i can multiply by 3 over 3 and then i'll have a 1 3 plus a 18 thirds which equals a 19 thirds well 19 thirds let's see that can be if you want to rewrite that as a you know mixed number you could you could also verify this as a decimals but i'm actually just going to leave this as a 19 thirds which again is approximately what six goes into 18 so it'd be six and two thirds no six six and one third right so just to kind of get an idea like 6.33 um you can kind of identify with there and again that like makes sense if like you look at this like my horizontal asymptote is six point is that six so therefore 0.33 would kind of make sense for my asymptote right so i'll just do a 0 comma 19 thirds for my y-intercept the other thing so now let's go and take a look at the domain range i want you to understand that just because i took this graph i shifted it you know to the right one unit but again did that change my domain no did that change my range no so if you take this graph and just shift it right horizontally that's actually not changing the domain of range because the only thing that's restricted on my parent graph is the range right the range is cannot be below a negative number it cannot be below this horizontal asymptote so what happens is when i shift this now graph up six units well the range has now changed from six to infinity so let's go and identify the domain here so the domain is unchanged it's negative infinity to infinity it's all real numbers and the range though is now going to be to its lowest value which is that 6 which is my asymptote and to infinity let's actually go ahead and move this information up here put over here so then everything is by next to each other there we go and then we'll just go ahead and call this the y-intercept so my y-intercept right is going to be what i say 0 comma 19 over 3. okay so my domain range you can see now the graph is only going down um as low as six here and then it's going up to if any whereas the graph going from left to right is expanding from negative infinity to positive infinity and there you go guys that's basically what i want you to be able to know for the graph is to be able to identify the transformations identify the y-intercept and identify the domain range okay so let's go and take a look at another example so we can go ahead and practice and you can see in this example we do not have the same type of function right because this has a base that is less than one so this is going to be a decay function so again remember like when we're looking into graphing it i want to be able to identify my transformations based on the paragraph so the paragraph in this case is actually not going to be a growth function right it's actually going to be the k so it's important that here is my decay function it's going to look something like this and again this is just the paragraph right so let's go and look at the transformations that we have and then kind of identify well how is that going to change this graph so one of the big important things here is we have this 2. now this 2 is positive so that's good but it's also going to be a vertical stretch of the graph which is actually going to change the y-intercept so be very careful when you recognize this and again we'll see how that works by just using a simple point to demonstrate that so again we're going to have a vertical stretch of two since i'm adding a five inside the function right it's as an exponent that's gonna be shifting it five units to the left and then i'm shifting it up one unit so up one unit okay so now if i just wanted to go ahead and simply grab this again what i can do is i can simply now again we're not trying to be exact here right but what i want to do is basically take this graph here and i'm just going to shift it five units to the left and then up one unit now to graph this we can basically apply the transformations but again we don't want to use this point here because this 2 is actually impacting it so the way i can explain this is let's just take a look at a quick little function let's say i have g of x times 2 raised to the one half x right so that graph looks something like this right but again watch out when i plug in x of 0 or g of 0 right look what we have we're going to get one half raised to the 0 power which is now going to give me a 2 times 1 right so what i want you to understand here is that's going to be equal to 2. so g of 0 is actually equal to 2. so what happens is whatever my scalar is that's actually going to tell me what my new y-intercept is so in this case i don't want to shift 0 comma 1. i actually want to shift the point 0 comma 2. and so what i'm going to do is i'm going to go left 5 units 1 two three four five and then up one more unit so i have zero two i'm gonna go over five and up one so this new point that i have on the graph is going to be at a negative five comma three right so it's important to kind of recognize that we also have our y in our sorry our horizontal asymptote which is just being shifted up one so therefore that's going to look like that and then we can also find the y intercept now again to find the y intercept we know that x has to equal 0. so we're going to go ahead and do this in the exact same case here so what i have here is 2 times a 1 half raised to the 0 plus 5 plus 1. 0 plus 5 is just going to be a 5. so therefore i have a 2 raised to the one-half to the fifth power plus 1. now one-half to the fifth power is going to be one to the fifth which is one and then two to the fifth which is going to be a 32 and now what i'm simply going to do is i can just multiply this 2 over 1 times 1 over 32 and then add it to 1. but again i want this one to have the same denominator as 32 so what i'm going to do is i'm going to rewrite 1 as 32 over 32. okay and now you guys can see that my y-intercept is going to be a 34 over 32 so just a little bit over one which again kind of makes sense right because i have a horizontal asymptote here at 1. so my y-intercept is going to be 0 comma a 34 over 32 which again dividing by 2 on the top and bottom could be reduced as a 17 over 16. and again that's going to be like that little point right there 0 comma a 17 over a 16. okay and then the graph we know is going to have that kind of pattern now it's also important it's approaching the asymptote right it's not crossing so it's important also for us to recognize our domain and range now notice in this example my domain again by these transformations did not change right it's still going to be negative infinity to infinity and my range is basically still the same but now the lowest value is going to be a positive one so my domain here is going to be the same so negative infinity to infinity and my range here is going to be from a let's see a 1 comma to infinity the only other time that the rank would change is if we have a reflection about the x-axis but i want you to understand since the domain is all real numbers it doesn't matter what transformation i do shift left right up down reflect x reflect wide it doesn't really matter it's not going to impact the domain now let's just go and do a quick little review of our logarithmic function so again when we're dealing with logs again we need to know what our function is we need to know our general function so log you know we have like log of x or also we could both be dealing with uh ln of x and it's important to recognize that for both these graphs it's going to take down the shape just like we had for the exponential function which is actually the inverse of the function so if you remember the exponential function of y equals b to the x and again this is helpful like whenever you like if you do forget it you know y equals b to the x that's just the equation form if you wrote it like this what i want you to understand here is this is just the inverse right so what we have here is this point is one comma zero it's just swapped so therefore the graph is going to you know take this kind of shape here and then again if we have a horizontal asymptote and this one well now for the logarithmic function it's going to be here now again just kind of like we had for the exponential function we have the base b it doesn't really matter what the base is it could be 10 it could be 5 or it could even be e the shape of the graph or the general shape of the graph is going to still be exactly the same obviously that is going to increase and decrease that's going to change as we're graphing it but the general idea of it's going to be the same now we're also going to have our transformations just like our exponential so i can say you know f x is equal to an a log of b of x minus c plus d again a is going to be our vertical stretch or compression and when a is less than zero it's going to be a reflection about the x-axis and then c and d are going to be our vertical shift so it's just important to recognize how those are going to be impacting our graph so let's just go and take a look at you know a quick little example of you know how to graph them and then how that's going to impact the domain and the range as well as how it's going to impact the graph so let's go and take a look now at just an example to see how i can identify the domain in the range okay so in this example uh we have ln at times x minus three plus one so again it's very important to understand the difference here the three is inside the function right that's our c so that's going to be our horizontal shift whereas the one is outside the function so that's going to be our vertical shift so it's very important to know the difference because when it's inside it's horizontal outside it's going to be vertical now there's no other stretch or compression or reflection that's going on here so in this case what we can look at is let's just go ahead and sketch the paragraph y equals ln of x and that's going to look something like this right and now again we know this point here is at one comma zero right so now let's just apply the transformations let's shift this graph three units to the right and then up one unit so we can go one two three and then let's just go up one unit and voila there's our new point but remember we also have a vertical asymptote so if we're shifting left or right that vertical asymptote also has to be shifted so now i'm going to go over three and units 1 2 3 and i'm going to draw a nice little vertical asymptote right here and then now i can just go ahead and sketch my graph now if you remember from exponential equations right my domain was unlimited right it could keep on going to left keep on going to the right the range though was limited to only positive numbers now look at over here when you look at the domain and range of this parent function the domain is now restricted right it's only restricted to positive values whereas my range is not restricted it looks like the graph is going to continue going down and continue going up so here let's just look at this new restriction it was for only positive numbers but now the vertical asymptote got shifted over three units so my domain in this case is going to be from three to infinity and again i'm going to use an open parenthesis because it's not an asymptote it's actually never going to achieve three and my range is going to be all real numbers all right let's go and take a look at one more example because i do want to address the transformations which we haven't really talked so much about so i want to be able to make sure we can address those as well okay so in this example we have a lot of negatives going on here but again it goes back to that inside and outside identity identification i want you to recognize that this is a negative and this is a negative both on the outside that means they're going to be vertical transformations now this is basically a negative 1 being multiplied by the log whenever you have multiplication by negative that's going to be reflection and again if it's vertical that means it needs to be reflected about the x-axis so therefore my first transformation is going to be reflect the x-axis now here i'm subtracting a 1 on the outside of the function so therefore that's going to be a vertical shift down one now the last point we have here is we have a couple things uh what we have here is we have this negative 2x well what exactly is this negative 2 doing right i talked about this in more graphing a logarithmic equation so it is important for you to recognize what happens when we do a b now we need to understand what is happening on the inside now remember the it whatever's on the inside is kind of like the opposite of what's going on the outside right so if the multiplying by negative on the outside is reflect the x-axis well what do you think multiplying by negative on the inside is reflecting the y axis if multiplying by 2 on the outside is going to be a vertical stretch of 2 well then multiplying on the inside is going to be a horizontal compression of 2. so that's exactly what's happened to the graph it's being horizontally compressed and is being reflected about the y-axis okay so now let's go and see how are we going to graph this and then identify the domain range and again we got to start out with what the function the parent graph is going to look like so that's just going to be a you know f of x equals a log of x and again that graph is going to look something like this it's going to have a y-intercept here at one comma zero and voila all right and again remember we have a vertical asymptote going on right there so let's see what happens now the first thing we're going to do is actually apply the restrictions from the from the inside first so the first thing i'm going to do is i'm going to reflect this about the y-axis so reflecting this about the y-axis i'm going to now produce this graph right so that's going to be a you know let's just say the negative 2x is being applied there and then it's saying well in addition to that now you need to reflect this about the x-axis so now the graph is going to be looking like this right so now it's a negative log of 2x and let's just call that a log of 2x and then last but not least is we need to shift it down one right so now the graph is being shifted down one so it's going to look something like that now it's important to understand that the actually the asymptote did not move right because all it did was i reflected the graph and again let's just kind of look at this one more time here from these more transformations so the original graph looks like this right now this graph is going to be shifted reflected about the y-axis is here then the shift about the x-axis didn't change anything and then it just got shifted down one so graph looks something like that and again still has my horizontal asymptote now what we want to do is we'll let's make sure we can identify the domain and range right so my domain is going to be all negative numbers to zero so negative infinity to 0 and my range is going to be again you know unrestricted so negative infinity to infinity there you go hopefully now we have a good foundation of the relationship between the exponential function and the logarithmic functions especially looking at them from a graphical approach because remember they're inverses of each other they undo each other you can see how to go from one to the other that's very important because for the rest of the problems in this chapter we want to make sure that we can relate our log method functions to our exponential and our exponential to our logarithmic and again that just kind of starts with like well how do we understand then a exponential equation in logarithmic form so again it's just really important to recognize like if i had 8 squared we know that answer is going to be 64 right but we can also understand this this is exponential right we have an exponent we can also understand this in logarithmic form and the way i like to explain this the best with the students is just remember base base base base you've got to remember this is your base right and that is going to be your exponent so when you have a logarithm logarithms also have a base so the base is always going to remain the same now that we have the base again we want to explain this equation but in logarithm form so this equation says 8 raised to the second power equals 64 which we know is true in this example if we want if we didn't know what the exponent was and we wanted to solve for it we could say 8 raised to what power equals 64. well that answer is going to be 2. so in a logarithmic equation what we're doing is we're looking for the exponent so that is going to be the answer we're going to be trying to solve so again just make sure you can kind of apply and understand the relationships a lot of times just remember bases are the same exponent is is what you're trying to solve for in logarithmic and the value is what you're trying to solve in exponential which is what we call the argument of the logarithm now this is really helpful in not only solving but also just evaluating logarithms so let's just work through a couple examples to make sure we really have a good understanding of evaluating logarithms okay so in these examples again what we're trying to do is we're trying to find the value of the logarithm and again a lot of students will get confused on these because logarithms can just be tricky so if you do get confused remember this is a this is an argument right so it's going to be equal to a value right and again if you can remember that it's equal to the value and what are you trying to find for the logarithm you're trying to find the value of the exponent so there basically this can be rewritten as 9 raised to what power equals 81. and that's how we read a logarithm right we say log base 9 of 81 but really what that means is 9 raised to what power equals 81. now hopefully in this example this one's easy enough for us i identify that x is equal to 2 right so we can just say that the log base 9 of 81 is equal to 2. now not always is the case is it always going to be you know that easy so in this case we have 81 raised to what number is 9. now hopefully with some experience you'll be able to get this one but again if you get stuck just set it equal to x right so we can say 81 raised to x equals 9. and remember solving exponential equations we're going to get to here in just a second we can rewrite 81 as a 9 squared equals 9 and therefore we can say 2x is equal to a1 and then you can divide by 2 divide by 2 x equals a one half in this example again we kind of have the same thing right set it equal to x so 9 raised to x equals a 1 9. well again 1 9 we can rewrite as a 9 to the negative first power so 9 to the x equals 9 to the negative first so x equals a negative 1. now in this example hopefully this one you can really kind of quickly see that 9 raised to what power equals 1. well again any number raised to the 0 power is always going to go 1. so we can find this answer just to be 0. now again i'm including x in this case you do want to just set the value equal to zero like i'm adding in an x just for explanation purposes in these examples now again when you have fractions again this is where a lot of times students will get confused so 1 9 you know raised to the x equals 81 and again just rewrite it so 9 to the negative first power right raised to the x equals a 9 squared well again here then we have a negative x equals 2 so x equals a negative 2. so that's the value of that function 9 raised to what power equals 9. that one's easy that's 1. 9 raised to what power equals 0. that's interesting think about it 9 raised to what number is equal to 0. now again remember what is an exponent an exponent means a repeated multiplication right 9 raised to the 0 power is 1. so is it possible in our real number system to raise 9 to a number to get 0 and not in our real number system so that is going to be it does not exist and the same thing here like 9 raised to what number equals a negative 9. well it doesn't matter how many times i multiply a 9 by a positive or a negative number it's never going to make a negative value because remember 9 to the negative first power that equals a 1 9. so again this is another example of a value that does not exist so whenever you get stuck when you're evaluating your logarithmic equations just remember to kind of think of them in exponential form to be able to better understand them so now that we've practiced a little bit more with logs it's time to kind of get into the properties of logarithms that we need to make sure we know okay so our three main properties is going to be the product the quotient and the power and again a quick way like a quick simulation for these is remember the rules of exponents basically going to be kind of like the exact same thing right and there's you can see that relationship that we talked about when we were first introducing these then the main thing we have here is whenever i have a log of a base you know whatever the base is when i have a product inside that argument i can rewrite that as two additional logarithms but remember these rules go backward go the other way as well if i have the logarithm plus another logarithm as long as they have the same base i can multiply the arguments and again that kind of also works with radicals if you remember like the the rules of radicals we have the same kind of idea for the quotient again when i have a log base b of m or n remember we can subtract them and again as long as we are subtracting two logarithms with the same base we can rewrite them as the as a quotient and then the power rule whenever we have a logarithm with the argument raised to a power we can rewrite that in front as multiplication so our properties of logarithms are really helpful for us to be able to condense and expand expressions which can be helpful for just simplifying expressions or solve an equation let's just go and take a look at two examples of at least condensing as well as expanding all right so in this example remember whenever we have parentheses the main thing we're going to do is work inside those parentheses first right go back to the order of operations so in this case um what i recognize is you always want to apply the power rule if you can first now this is in arguments here ln of 4x but i can go ahead and rewrite that as up front here as a power so 3 ln of 6 minus let's rewrite this as the ln of 4x raised to the 3 halves plus a ln of 5. and again that's in the argument but now you notice that ln both have a base e i can rewrite these as the product now again this is going to be raised to a power which again is going to be the same thing as the square root of 4 x cubed so therefore i can go ahead and rewrite that as a radical so therefore i have a 3 l and a 6. then i'll just go ahead and rearrange this as a 5 times in front so therefore this would be an ln here of a 5 and then this is going to be the square root of a four x and that's gonna be a quantity to the third power okay so now i recognize um i forgot to do the power rule here so that can be written as a ln of six cubed right which is going to be a 216 so ln of 216 minus here a ln of 5 times the square root of 4x cubed okay now again notice that we have both ln's right they're both you know base e so now we can see we're subtracting so therefore that's going to be looking for the difference here so then my final condensed version here is going to be a 216 divided by a 5 square root of 4x sorry raised to the third power okay and you can see that what we did is when we're condensing we condense them all the way down to one single logarithm now in the second example what we have is we have one single logarithm what we now want to be able to do is expand this to multiple logarithms so the first thing we're going to do is i'm going to rewrite this radical as a rational power just like here how we wrote the rational power as a radical now we're just going to kind of go in the reverse operation so since everything is under that quantity of the radical i want to be able to write that as a rational power so what i'll do is say log and now notice the power rule i can take all my powers and rewrite them down in front and now what you recognize is each of these values inside of my argument is separated by multiplication now again this one half is still multiplying by this whole logarithm but what i want you to be able to see is now i can separate each of these into their own logarithm and i just want to make sure i'm using parentheses again to represent that all of those functions are going to be multiplied by one half okay so there you go ladies and gentlemen you can see now by expanding i had to rewrite the example and using multiple logarithms but the process is the same when you're condensing you're trying to go down to one example whereas expanding we're trying to expand them using multiple and we're using the same properties of logarithms but just in a different order now we already did a quick little view review of solving exponential equations when we were evaluating logarithms but it's also important when we have a little bit more complicated examples so you know just remember like if i have 3 squared is equal to you know 3 to the x like how do we solve this well again we don't really need to do much right whenever our bases are the same we know that the powers have to be the same as well so x is going to equal to 2. so there's a couple different ways we can use this so we always want to first look to see if we can solve using what we call the one-to-one property so in this first example you can see that my bases are not the same but can i rewrite these as the same base and one thing i recognize is 9 and 27 are both powers of 3. so i can rewrite 9 as going to be a 3 squared right still raised to the w minus 8 and then 1 over 27 i can rewrite as a negative 3 right because remember negative you're just going to reciprocate and again that's still going to be written to the 2w now remember the power rule of exponents whenever you're whenever you take an exponent raise it to another exponent you're actually multiplying and again since my bases are the same i can apply this one to one property so therefore this is going to be 2 times a w minus 8 is equal to a negative 3 times a 2w okay now i can just go and apply distributive property notice how i inserted the parenthesis here that's very important so therefore i have a 2w minus a 16 equals a negative 6w now i just want to get my variables to the exact same side so i'll add a 6w to both sides and i get 8w minus 16 equals 0 add a 16 to the other side so 8w equals 16 divided by 8 divided by 8 w equals a 2. what about when you only have one exponent right and you don't have you can't use the one to one property so in this example what we're going to want to now do is use what we call our inverse operations we want to isolate our exponential equation so in this example what i'll do is i'll add 11 to both sides and i'll have a 2 times a 3 raised to the 4y is going to equal what a 72 then i'll divide by 2 on both sides and i get a 3 to the 4y is equal to a 36 now you might be saying well how do i solve for this now again this goes back to our solving our logarithmic equations right i don't know how i can solve for y but again when we want to solve for exponent what do we do we look for our logarithmic equations so what i'm going to do is i'm going to convert this to a logarithm and again remember the base is the same so i have a log base 3 and again of 36 is now going to equal to a 4y so then i'll divide by 4 on both sides and i can say y is equal to a 1 4 log of 3 36 now how are you going to how are you going to figure out the 3 raised to what power equals 36 because it's not going to be an integer so again remember when we have a logarithm of different bases we can use what we call the change of base formula so what we'll do is i'll rewrite this as a 1 4 which again is the same thing as dividing by 4 as a log base 10 or base e or any base really of 36 over a log of 3. now that's what i'm going to plug into my calculator and let's go and see what our answer is and just using a little rounding to my nearest thousands i get 1.305 now on this last example for some reason students just get confused as heck when we have e it's really going to be the last same example that we did in the previous one all you're going to do is if there's only one exponent is you know isolate your exponent so in this case i will have a e raised to the a plus 1 is equal to a 5. now again i can't undo this a plus 1. i have to get it outside of my exponent before i can actually sulfurous now in this example we rewrote this as a logarithm which you could also do in this case i can also just take the logarithm of both sides with the same base so therefore i could have ln of e to the a plus 1 equals the ln of 5. and again that's going to be base e now again remember the rules of exponents say as long as the same base of the argument then that's just going to equal my power which is a plus 1 equals a ln to the 5. then i'll subtract 1 on both sides and again you're going to want to use your calculator here so therefore this is going to be a ln of 5 minus 1 which again using my calculator is going to be approximately 0.609 now for solving logarithmic equations we're going to want to use the same process we're going to look into using inverse operations as well as a one-to-one property but we also have to be very careful to check our solution because we can also have an example of extraneous solutions and again remember extraneous solutions is just when we have a solution for a simplified version but not the original equation now in this first example what i want you to recognize is i have a logarithm equal to a logarithm and whenever you have a log equal to a log the best easiest fast thing to do is just like we did for exponents is set the arguments equal to each other so now i can just rewrite this as x squared minus x equals a 6x plus 18. so guess what we're doing a logarithmic equation and we quickly just now converted this to a quadratic now when solving quadratics remember we want to get everything to the same side and set it equal to 0. so to do that i'll subtract a 6x on both sides and subtracting 18 on both sides therefore now i'll get a x squared minus 7x minus 18 equals zero and now i can just go ahead and factor this you know what two numbers multiply to give me negative 18 but then add to give me a negative seven so since this is a negative one's positive and one's negative my nine is gonna have to be negative and then plus two it's going to be a negative seven here that's what i'm getting so i'm gonna say x minus nine times an x plus two is equal to zero now i can just use the zero product property set them both equal to zero and therefore x equals nine and x equals negative two now again remember what i said we got to make sure we're plugging these back in to make sure that these are going to work so basically we since the log is equal to log we just want to make sure the arguments are going to work on both sides so this would be 9 squared which would be 81 minus another 9 which is going to be a 72. so we want to make sure this is going to give us a 72 9 times 6 is 54 plus 18 which again is going to be a 72. what about negative 2 so negative 2 squared is 4 minus a negative 2 is going to be 6 that works negative 2 times 6 is going to be negative 12 plus 18 is 6. so both of these solutions are going to work now in this example one of the biggest mistakes students make is they see this and they're like oh let's just go ahead and cancel out the logs and no we don't want to do that in this example you can only do that when you have a logarithm equal to another logarithm so what we're going to want to do is use our rules of logarithms to simplify this so i see there's a 2 in front so i'm going to bring that up as a power and then these i can condense these using the quotient rule so now i have an ln of y plus 5 quantity squared equals ln of 20 over 5 right which actually can be reduced down to a 4. however notice i have a logarithm equal to a logarithm it doesn't matter if it's a natural log or not right so i can now just say y plus 5 quantity squared equals 4. now let's go ahead and simplify this right so y plus 5 quantity squared is going to be a y squared plus 10 y plus 25 equals 4. and again this is another quadratic so again we're going to want to set this equal to zero so i'll subtract it over so i have a y squared plus 10 y is plus a 21 equals zero now this factored form is going to be well two numbers multiply give you 21 add to give you 10. well 7 and 3 right y plus 7 times y plus 3 they both have to be positive equal 0 and then i can say y plus 7 equals 0 and y plus 3 equals 0 y equals negative 7 y equals negative 3. okay so now we've got to go back and check our solutions we don't want to check this solution we already know this is true we want to make sure these solutions are going to satisfy the original equation and already i can see that negative 7 does not work because look what happens when you take negative 7 you plug it into the equation negative 7 plus 5 is a negative 2. you cannot take the logarithm of it with a negative argument so therefore this is going to be what we call an extraneous solutions now let's just go and take a negative 3 here just real quick so negative 3 plus 5 is a 2 and then 2 raised to the second power would be 4 right and we know that's going to be ln of 4 equals ln 4. so therefore this solution is going to work now in this last example you can see that we only have one logarithmic equation we're not going to look into using the one to one property what we are going to want to do though is isolate our logarithm so we're going to do that now by adding a 14 and dividing by 2 on both sides now what we're trying to do is we're trying to solve for m which is inside of the argument so there's a couple ways we can do this what we can simply do is just rewrite this in exponential form right if i rewrite this in exponential form i'm going to have a 2 raised to the 7th power equals a 9m plus 2. now i just need to know what two to the seventh power is which is going to be a hundred and twenty eight equals nine m plus two and then i'll subtract two on both sides 126 equals a nine m uh divide by nine on both sides and m is going to equal a 14. again we want to make sure this is not going to make our argument negative which indeed it does not so we are good to go another way you could also look at this is also using your rules of exponents when you have the log base 2 of a 9m plus 2 equals 7 you can also raise both sides with a base two so then that would just leave you a two m plus two equals a two to the seventh which is the exact same thing we achieved there it really just kind of depends on how familiar you are with the rules of your logarithms when you're taking the test and the next thing i just want to cover is going to be simple and compound interest it's just very important to make sure we recognize these formulas and maybe and maybe actually even remember these or practice them at least inputting some different values so when we're dealing with simple interest remember that's just going to be the total amount of interest based on the time and the interest rate off of your principal if interest you earn is not going to be growing on itself if you want to find the simple interest we're just going to take our principal times 1 plus r and then raise it to t where t represents the number of years now compound interest is where you earn amount of interest but then every single compounding period typically daily monthly yearly it's going to grow it's going to take that amount and compute the new interest based on what you previously earned so we want to find compound interest we have a equals p times 1 plus r divided by n so you're going to take the number of compounding periods you're going to take your interest rate divided by the number of compound periods and then you're going to raise it to an n times t so this n is going to be very important now let's just take a look at two different examples let's say you invest five thousand dollars so we're going to say p equals you know five thousand dollars and let's just take a look at two things now your first way you could do it is let's say r equals a two point three percent now this let's just say this is like in a bond program like with the in the government right and they're going to say we're going to give you this interest rate and let's say t is going to equal 15 years all right because you don't want to lose your money and this is a you know it's with tied to government it's going to be a very safe bet so they say all right in 15 years at 2.3 you know basically there's zero chance that you're going to lose your money and you're gonna invest five thousand dollars how much money will you have at the end of 15 years so it's very important when you're importing this to not use 2.3 but use .023 so now all i'm simply going to do is i'm going to go ahead and simplify that to a 5000 times 1.023 and raise it to the 15th power now i'm going to raise this to the 15th power first and then multiply by five thousand and when i do that i have seven thousand thirty two dollars and forty two cents which again is not too bad right i mean if you didn't do anything if you didn't have to do anything you didn't have to work for that you kind of made an extra two thousand dollars pretty good but what about if we have something where we have an end where it's going to be compounded even though we calculate the stock market as a compounded on a yearly front let's just look at this and pretend that we get our gains every single month and we're going to have things compounded on a monthly term so let's say you take the same amount of money now again the stock market you have much more risk companies could go bankrupt and you could lose all of your money however because of that risk let's say you're going to earn a much higher percentage let's say you're going to earn 8.1 percent so you're going to take the same amount of money 5 000. let's have the rate is going to be an 8.1 percent we're still going to use 15 years but now we're going to have our compounded period is going to be monthly so therefore per year that's going to be 12. now let's just go and plug in these values into our formula now we'll just go ahead and simplify this so a equals you know a 5 000 and then we'll have a .081 divided by 12 plus 1 which is going to be a 1.006 and that's going to be raised to the 180th power so i'm going to take this raise it to the 180th power and then multiply that by 5000 and just by doing that ladies and gentlemen i have 16 782 dollars and 82 cents so that's a huge difference a huge difference right same year compounding period is a little bit different right but even if you did yearly it would still be a large it would still be a large value but the interest rate changed a lot and also that this value was compounding you actually not just earned 12 000 you actually tripled your money in 15 years pretty cool so just remember whenever you're doing the compound or simple interest make sure you understand if it is simple or if it is compounding and again compound you're going to have a compounding period and again basically it's just going to be plugging them in and then simplifying hopefully this video was helpful for you preparing for or reviewing exponential and logarithmic equations if you want more examples then check out the playlist and resources i have down below otherwise i look forward to seeing you guys on the next video cheers