here are the top 10 most important things you need to know about exponential functions must know number one the graph and properties of an exponential function the general format of an exponential function looks like this y = a * B to the^ of X in this equation X and Y are the variables that have an exponential relationship A and B are constants let me tell you a bit more about A and B A is the initial value of the function and graphically that would show up as the Y intercept at any Y intercept we know that the x value is zero so in the equation if you make x0 this power would equal 1 leaving y to just equal a and it's also worth noting that a can be any real number now let's look at B notice that b is the base of the power and there are actually a couple restrictions on what the value of B can be the B value has to be greater than zero so that the value of the function is always a real number and the B value can't be equal to one so that the function isn't just a constant function now I want to show you what the graph of an exponential relationship looks like so I will do that by showing you the graph of a very simple exponential function f ofx is equal to 1 * 2 to the^ of X which could be Rewritten as just 2 to the^ of X but but I wanted to write this one here just so you could see that that's going to end up being the Y intercept or the initial value of the function before I sketch its graph let's make a quick table of values I'll choose X values between -3 and 3 so we get a section of the graph that is a good representation of what an exponential relationship looks like and now I have to take each of these X values sub them into my f ofx function for x and then calculate y when looking at this table notice that as the X values increase by one there's not a common difference that the Y values increased by but there is a common ratio to get from one y value to the next you would just have to multiply by two and hopefully you can see the connection of why the common ratio of two is occurring it's because the base of our power in our function is two this initial value is just being repeatedly multiplied by two in x amount of times and then if I plot all of these points on my graph and then connect them with a curve I can see what the en shape of an exponential relationship looks like but when I drew this graph I left at one key component of it notice as the function is moving to the right the Y values are doubling from 1 to 2 to 4 to 8 to 16 and so on but if I were to think about what's happening to the Y values as we move to the left they go from 8 to 4 to 2 to 1 to a half to a quarter to an eth they keep getting cut in half are those y values ever going to get down to zero no it doesn't matter how many times I cut this value of .125 and half I'm still going to be left with something no matter how infinitely small that's why there's a horizontal ASM toote at the line Y equals 0 which is right on the xaxis so let me communicate using a limit why the horizontal ASM toote is at y equals 0 it's because the limit of our 2 to the X function as X approaches negative Infinity is equal to zero it means this function as X goes to netive Infinity the Y values are going to get closer and closer and closer to zero and the last main property would be this function's domain and range the domain means the set of X values the function can take so what can we put as the input for X into our 2 the X function and be able to get an answer well anything and graphically we can see that because this function goes forever to the right and also forever to the left so our domain would be any value between negative Infinity and infinity the range however means what's the set of Y values the function could output so what could we get 2 to the^ of x equal to well notice the function is always above our horizontal ASM toote so the function is always going to have a y value that's bigger than zero so the range for this function would be from0 to infinity and notice I put a round bracket at zero because we don't want to include zero the function can never actually be zero it's always greater than than zero must know number two exponential growth vers exponential decay we'll start with exponential growth exponential growth just means we have an exponential function that is increasing and a function that is increasing means that as X increases y also increases that just means in general the function is going up to the right and exponential decay means we have an exponential function that is decreasing and a decreasing function is a function that as X increases Y is decreasing now remember the general format of an exponential function looks like this y = a * B to the power of x a is the initial value and that initial value gets repeatedly multiplied by the base of the power x times depending on whether A and B are positive or negative we could get a different shaped exponential function and in fact there are four different shapes there are two shapes that represent exponential growth and two shapes that would represent an exponential decay let's look at this first graph notice it starts above the xais so the a value must be positive and as we move to the right the Y values are increasing so the initial value the a value must be being multiplied by a number that's bigger than one to make it increase so the B value must be bigger than one but for this graph its initial value it's a value must be negative because it's below the x- AIS and for that function's y- value to increase as we move to the right the initial value the a value must be being multiplied by a number between 0 and 1 to make it get closer to zero if we look at these decreasing functions the first one a is positive but the Y values are decreasing so that a value must be being multiplied by a number between 0 and 1 and this last shape notice its initial value is negative and then the Y values are decreasing as we move to the right to make them go further in the negative direction that negative a value must be being multiplied by a b value that is bigger than one repeatedly to make it go further in the negative Direction so those are the four main shapes let me do a quick example of a problem that involves both exponential growth and exponential decay let's start with an example about exponential growth this example says if you start with one cent and it doubles every day how much will you have after 30 days I know the scenario can be modeled with an exponential function because we have an initial value that is repeatedly being multiplied by two so I can write this in an exponential format y = a * B to X in this format a is the initial value which in this question is 1 cent B is the base of the power which for this question has to be two because the initial values repeatedly being multiplied by two and X stands for the number of times the initial value gets multiplied by that growth factor of two we know it doubles every day so after 30 days it's going to double 30 times so our exponent on the base of our power is 30 if I sub these values into my exponential function I'll be able to solve for y which is how much money I have after 30 days when evaluating this you're going to see the power of exponential growth we're starting with a very small amount only 1 cent but we're doubling it 30 times which is going to make that one cent grow to over $10 million it's going to grow to 10,737 41824 let's now do an exponential decay example this example says that uranium has a half life of 2 years how much of a 1,000 MGR sample will be left after 10 years now in this example this word halflife just means the amount of time it takes for a sample to be cut in half so every 2 years this 1,00 milligram sample it's going to get cut in half well since there's a repeated multiplication by a half we can model this with an exponential function again I'll just have to figure out what my parameters a b and X are and then I could solve for y which is the value of the sample after 10 years so my initial value is what I start with a we know that we start with a 1,000 mgram sample B is the base of the power that means what are we multiplying the initial value by repeatedly well we're multiplying it by a half every 2 years so the base of our power is a half and X the exponent on the base is the number of times that we need to multiply the 1,00 by a half well in 10 years how many times is it going to be cut in half if it gets cut in half once every 2 years we can do a little calculation we can do the total amount of time and divide that by the half life of the sample which is 2 and of course 10 / 2 is 5 I'll write the simplified version when I write my equation that means this 1,000 mgram sample is going to be multiplied by a half five times and when you do that you'll be left with 31.25 mg of uranium must know number three how to find the equation of an exponential relationship from its graph I can tell from the shape of the graph this looks like an exponential function I can Define the relationship between the X and Y values of every point that makes up this function using an exponential function of the format y = a * B to ^ x I now just have to solve for what values of A and B give us the graph of this exponential function to solve for two variables I need two equations if I look at the graph there are two points labeled for me each of those points has an X and A Y y value now with each of these points I will generate two separate equations by subbing their X and Y values into this General format of an exponential function so using my first point my equation would be 5 = a * B to the^ of 0.5 and using my second point the equation will be 10 = a * B to the^ of 1 and now I can use the method of substitution to solve this equation what I'll do is in this equation I'll isolate the variable a by dividing both sides of the equation by this power of B and then I can substitute that value of a into my other equation for a and then be able to solve for b now these powers of B are being divided using exponent rules I know that what I can do is I can keep that power and then subtract their exponents 1 minus a half is a half I'll then divide both sides by five to get rid of that five and then to fully isolate B I'll have to take care of that exponent of a half I could get rid of that by squaring both sides of this equation and that would tell me that 4 equals B to the^ of 1 which is just B and finally we can take that answer for B sub it in here which already has a isolated 4 to ^ of 0.5 is just theare < TK of 4 which is 2 so we have an a value of 5/2 which could be written as a decimal as 2.5 and and then we can see on the graph remember the a value represents the initial value it looks like the Y intercept is at 2.5 and B is the base of the power which in this case is four so that tells you what the initial value is repeatedly multiplied by every time X increases by 1 so if the initial value is at 2.5 when X increases by 1 the Y value should be multiplied by 4 which it is 2.5 * 4 is equal to 10 so the correct equation of this relationship is y = 2.5 * 4 to the power of X must know number four transformations of an exponential function the exponential function y equals B to the^ of X can be transformed with four different parameters a K D and C let me start by making a list and showing you a demonstration of how each of those parameters can transform an exponential function let's start with a the value we multiply the power by if a is bigger than one or less than negative 1 it causes a vertical stretch which I'll write as vs by a factor of which I'll write as bfo the absolute value of a but if a is between 1 and one it actually causes a vertical compression by a factor of the absolute value of a and if a is negative so if it's less than zero there's going to be a vertical reflection across the x-axis the other type of vertical transformation is caused by the C value that's added after the power if C is greater than zero the function shifts up and if C is less than zero the function shifts down and now let's talk about the horizontal Transformations which up in the exponent K and D if K is bigger than 1 or less than 1 there's actually a horizontal compression by a factor of 1 / the absolute value of K and if K is between 1 and one it's actually a horizontal stretch by a factor of one over the absolute value of K and if K is negative so if it's less than zero there's a horizontal reflection across the y- AIS and lastly the D value is what we are subtracting from X if D is positive it shifts the function to the right but if D is negative it shifts the function to the left let me make a little bit of room here so we can do an example where we graph a transformed exponential function we are asked to graph this transformed exponential function it's a transformed 1/2 to the^ of X function we have these three parameters which are causing Transformations the number in front of the power is the a value so the a value of -2 is going to cause a vertical stretch by a factor of two and also cause a vertical reflection to do those two Transformations I'd have to multiply the Y values by -2 the D value is what we subtract from X in the exponent we are subtracting a three which means the DV value is three a positive D value shifts the function to the right so is going to shift right three units and we could accomplish that algebraically by adding three to all of the X values and lastly after the power we're subtracting 2 which means the C value is -2 which based on this criteria tells us the function will be shifting down two units and we could do that just by subtracting two from all of the Y values of the points that make up the function so now all we have to do is do these transformations to a set of key points for what we call the base function 1 12 to the^ of x so I'll make a table of values for a half to the power of X I'll choose X values just between -2 and 2 and then calculate the Y values by subbing my X values into my function for x and now I'll apply my transformations to these points and now I'll make a table of values for my transformed G ofx function in the table I'm going to communicate these Transformations that I'm doing to my X and Y values to the X values I just have to add three and to the Y values I have to multiply them by -2 and subtract two so I will do these transformations to these points to each of these X values I have to add three and to each of these y values when I sub them in here I have to multiply them by -2 and subtract two and then I can just use these points to get a good graph of my G ofx function so I'll plot all the points my last point doesn't fit on my graph but that's okay but I do have to add where the horizontal ASM toote is remember the horizontal ASM toote of an exponential function starts at the x axis but the C value of the function in this case -2 will shift the horizontal ASM toote down two units so the horizontal ASM toote of this function is at y = -2 and then I'll connect the points on my function with a smooth curve showing that it's going to be approaching that horizontal ASM toote as X approaches Infinity must know number five inverse of an exponential function let's say we have the exponential function FX equal B to the^ of X then the opposite of that function which mathematically we call the inverse and we have a notation for the inverse that looks like this where this NE -1 that looks like it's in the exponent doesn't actually act like an exponent this does not mean 1 over FX it doesn't mean the reciprocal it means the opposite of f x and the opposite of B to the^ of X is log base B of X but what is this log function what does that mean well just like division is the opposite of multiplication a log function is the opposite of an exponential function it undoes exponentiation or another way of saying that would be that a log function finds the exponent to which the base must be raised to to make it equal to the argument in that definition I use two terms that you have to know base and argument the base of a logarithm is here it's a subscript B and the argument in this case is the X and what this means is that if we have a log function y equal log base B of X the answer to this logarithm is the exponent that would go on the base of the logarithm to make it equal to the arent of the logarithm so I could rewrite this in an equivalent way in exponential form by writing that b to the power of Y is equal to X let's see if you understand logarithms by doing a couple quick examples the first example just says to evaluate a couple logarithms if I want to evaluate a logarithm I just need to figure out what exponent goes on the base of the logarithm to make it equal to the argument well I know two the^ of 3 is 8 so the answer to this logarithm is 3 and then for this one I'll be looking for what is the exponent that goes on 3 to make it equal to 9 well I know 3^ s is 9 so the answer to that logarithm is 2 and let's actually evaluate one more what if we have log of 100 I don't see a base on this logarithm which means it's the common logarithm that has a base of 10 so what exponent goes on 10 to make it equal 100 is 2 10 SAR is 100 and now let's do one more example where we can see when a logarithm is going to be useful this example just says to solve this equation for x 5 to what exponent equals 20 to get an approximate value for it we can make use of the logarithmic function we can take the exponential form that we're given and we can rewrite it in an equivalent logarithmic form if I'm looking for the exponent that goes on five to get 20 that exponent I can get an approximate value for it by doing log base 5 of 20 and then you could use a calculator to get an approximate value for this and you would get about 1.86 must know number six solving exponential equations let me give you two exponential equations that we'll solve both of these equations look similar we have a power equal to a power and our unknown X is in the exponent that's why these are called exponential equations to solve these two equations we'll use two different strategies the equation on the left it would be easy to rewrite those Powers so that they have the same base when I look at the bases of the powers 9 and 27 I recognize that those are both powers of three I could rewrite the 9 as a 3^ squar and I could rewrite the 27 as a 3 cubed and then on both sides of this equation I could simplify using exponent rules on the left I have a power of a power the exponent rule tells me that I can multiply the exponents together so this becomes 3 to the power of 2 * x + 1 and don't forget that 2 will multiply by both the X and the 1 so it becomes 2x + 2 and then same thing on the right I can multiply 3 by 2x - 5 to get a new exponent of 6x - 15 and now that I have a power equal to a power where the bases of the power are equal to each other I know the equation could only be true if the exponents were also equal to each other so we can now set the exponents equal to each other and solve for x using inverse operations I'll bring my variable terms to the right and my constant terms to the left and then collecting my like terms I have 17 = 4X and if I divide the four to the other side I figure out that x = 17 over 4 which as a decimal is 4.25 now let's look at the second equation it would not be easy to rewrite five and seven as Powers with the same base so instead we'll use a different strategy we'll make use of the log function we'll start by taking the logarithm of both sides of the equation and then using some log rules we'll be able to solve this equation and if you're not familiar with rules for logarithmic functions make sure you check out my top 10 video on logarithms as well one of the rules you can use for logarithms is the power rule of logarithms that tells you you're allowed to take the exponent of the argument and write it as the coefficient of the logarithm so we can do that to the logarithms on both sides of the equation and then we expand both sides of the equation so do x * Log 5 + 3 * Log 5 on the left and on the right 2x * log 7 - 1 * log 7 I want to get the terms that have a variable on the same side of the equation so I'll get my variable terms to the right side of the equation and my constant terms to the left side of the equation and then since both terms on the right both have an X we could common factor that X from both terms and then to isolate X I would just have to divide this constant value to the other side of the equation now at this point you could type this expression on a calculator to get an approximate answer for X but if we wanted an exact answer for X we could use some logarithm rules to simplify this this three could be moved as the exponent on the argument of the logarithm making it log of 5 cub which is log of 125 and in the denominator same thing with this two I could write it as the exponent on the seven and 7^ S is 49 and now I can use the product and quotient rule of logarithms to simplify further when adding Logs with the same base we can write them as a single log by multiplying the arguments 7 * 125 is 875 and when subtracting Logs with the same base you can write them as a single log by dividing the arguments so that would become log of 49 / 5 and and then we could combine those into a single logarithm using the change of Base formula but I think I'll leave it like that and then also just write the approximate value of this which is about 2.97 must know number seven exponential equations of quadratic form well what's an exponential equation of quadratic form look like let me show you the general structure in this equation a b c and d are all constant values and our unknown our variable is X notice in this equation I have two powers that have the same base the base of both of those Powers is B and the exponents one of them is X and the other one is 2 * X when you have this structure of an exponential equation we can solve it by first letting U equal b ^ of X and then once you do that you'll have a quadratic equation that you can solve and once you have those solutions for you you can back sub to solve for x so let me show you what that looks like with an example now in this example I have two powers that have the same base their bases are both five and the exponents one of them is X the other one is 2x so I know that I can use these steps and to make it clear as to what I'm going to be doing I'm actually going to start by rewriting this power instead of writing it as 5 to the^ of 2x using an exponent rule I know I could rewrite this as 5 to the power of x^ s and now notice here and here I have 5 to the^ of X so what I'm going to do is I'm going to let U equal 5^ x and when I replace both of these with u my equation becomes U ^2 + 8 u - 20 = 0 which looks just like a simple quadratic equation to solve I could even solve it by factoring just find numbers who have a product of -20 and a sum of 8 and the numbers that satisfy that product and sum are 10 and -2 so I know this would factor to U + 10 * u - 2 and this product would be zero if either of the factors were zero this Factor would be zero If U is -10 and this Factor would be zero If U is 2 but remember we're not actually trying to solve for you in the original equation our unknown is X so we have to now back sub these answers into our let statement each of these u values actually equal 5 ^ of X so I'll make two equations I know 5 ^ x = -10 and 5 ^ of x = 2 and then I'll solve both of these equations this equation I'm wondering what exponent goes on X to get -10 well no matter how many times you multiply five by itself it's not going to become negative the range of this function is between Z and infinity it's always above zero so there's no way it could be -10 so there are no real solutions to that first equation but this second equation there is an exponent we can put on 5 to make it equal 2 and we could find that exponent using a logarithm if we want to know what exponent goes on five to get two I know that log base 5 of two would solve for that exact exponent so X would equal log base 5 of 2 and then you could evaluate that to get an approximate value and you would figure out that X is approximately 0.431 must know number eight compound interest this is one of the main applications of exponential functions is the ability to calculate compound interest and to calculate compound interest we can follow this formula where in this formula a stands for the future amount P stands for the principal amount which is just the amount you start with i stands for the annual interest rate n stands for the number of times interest is compounded each year which we call the number of compounding periods and T is the number of years so basically how this formula works if we know we have a principal amount that we invest in an account for T years at a given annual interest rate and we know that that interest is compounded n times each year we can use this formula to find the future amount of the value of the money in that account so let's do an example you invest $100 in an account that pays 5% interest per year how much will you have after 10 years if a the interest is compounded annually or B the interest is compounded monthly let's start with part A if we go on to calculate the amount in the account after 10 years I can find the amount after 10 years by taking my initial investment of $100 and multiplying it by 1 plus the annual interest rate which is 5% now when we sub that into our formula for the interest rate we have to sub it in as a decimal value so 5% means 5 out of 100 so 5 / 100 is 0.05 and then I have to divide that by n the number of compounding periods in a year well it's only compounded annually which means it's only compounded once a year so I'll divide that by one and then my exponent is the number of compounding periods times the number of years the money is invested for well the number of compounding periods in a year is 1 and it's being invested for 10 years and then it's just a matter of evaluating if you evaluate this you would figure out that your $100 has increased up to $62.89 but what would happen if we compounded this interest monthly instead of annually well a lot of the equation going to look the same except for the number of compounding periods in a year it's no longer just going to be one it's going to be 12 so I'll rewrite that same equation but change n to 12 so if we compare this to part A in part A we got the whole 5% interest 10 times but now in Part B we're only getting a 12th of the 5% interest but we're getting that 120 times so you're getting a 12th of the interest but you're getting it 12 times more times now that sounds like that should be the same thing but you have to take into account every time you get interest added to your principal amount your next interest calculation is going to be interest paid on top of the interest you've already earned which is going to mean that the higher the number of compounding periods the higher the value is going to become and in this case the amount after 10 years would be slightly higher it would be up to $164 and. 70 cents and as you keep increasing the number of compounding periods this value is going to keep getting higher but what would happen if this value approached Infinity that's going to lead us into our next section must know number nine the natural exponential function let me start by giving you the definition of the natural exponential function the natural exponential function is the function e to the power of X and what's the value of e to the power of X well the definition for that is that it's the limit as n approaches Infinity of 1 + x/ n to the power of n now this formula looks really similar to our compound interest formula and to help you understand the value of e we can think of it in terms of compound interest so let's say we were trying to evaluate e to the^ of 1 well we would just write that as e and we would calculate it by doing the limit as n approaches Infinity of 1 + 1 / n to the^ of n right I replaced x with one and I'm actually going to add another number in here let me add a number in front of this power let me just put a one there to keep it equivalent but so that you can notice that we can think of this one as our principal investment and we're getting paid an annual interest rate of 100% right 100% would be one as a decimal value and let's say we're just investing this $1 at a 100% interest for only one year and from the last section we know that as we we increase the number of compounding periods that's going to increase the value of this power but what be the limit of this value as n approaches Infinity this value is not going to go to Infinity it's going to go to an irrational number that we call E which stands for Oiler number and I'll write down the approximate value of e for you right here now this value e is what we call Oilers number it keeps on going forever with no patterns or repeats so it can't be expressed as a fraction which means that e is an irrational number and another thing I want to note is that when we make the base of a logarithm e we call it the natural logarithm and a notation for that is Ln of X the natural logarithm of X and now let me quickly show you what the graph of e to the X looks like and do an example that involves e to the^ of X let's start with the graph so for the function y = e^ x let me make a quick t of values and then do a rough sketch I'll just choose a few X values -1 0 and 1 calculate the Y values by subbing these in for x and if I to sketch this function's graph just like all other exponential functions there's going to be a horizontal ASM toote at yals Z and then if I were to roughly plot these points and connect them I would get this shape and now let's do an example involving the natural exponential function let's say we wanted to solve the equation 20 = = 3 * e^ X my unknown is in the exponent so this is an exponential equation I'll start by isolating That Power by dividing the three to the other side and then if I want to know what exponent goes on E to make it equal to 20 over3 I know I could calculate that exponent by doing log base E of 20 over3 and then notice what I have log base e is what we call the natural logarithm so I could rewrite this as the natural logarithm of 20 over3 and then you could get an approximate value for that using your calculator it would be 1.89 7 must know number 10 the derivative of exponential functions let's say we wanted the derivative with respect to X of the exponential function B to the^ of X the derivative rule for this tells me that the derivative of B to the x would equal b^ x * the natural logarithm of the base of the power B so you just keep the power exactly as it is then multiply by the natural logarithm of the base of the power let's do two examples where we would need this rule let's find the derivative of 3 ^ x and then let's do another example where we calculate the derivative of the natural exponential function e to the^ of X well for the first example example if I want to differentiate 3 to the^ of X the derivative rule tells me to keep the power exactly as it is and then multiply by the natural logarithm of the base of the power so multiply by the natural logarithm of three that's it there's the derivative of that function and following those same rules for the second example I can keep the power exactly as it is and then multiply by the natural logarithm of the base of the power and the base of the power is e and then what is the natural logarithm of e well the natural logarithm of e remember the natural logarithm just means we have log base e of e and if you understand logarithms that expression means what exponent can you put on E to make it equal to e well the answer to that would just be one so this whole expression right here the natural logarithm of e is actually just 1 which means the value of this is e to x * 1 which is just e ^ x that's the most fascinating thing about the natural exponential function its derivative is itself and now let's make it a little bit more difficult let's say we're finding the derivative with respect to X of an exponential function where the exponent isn't just X it's a function of X so for that you would need the chain Rule and that would tell you to keep the power exactly as it is multiply by the natural logarithm of the base but then then based on chain rule we would then have to multiply this by the derivative of the function in the exponent so we would then multiply by F Prime of X so we'll do one example of this where we find the derivative of the function 5 to^ of 3x + 1 the derivative rule tells me to keep the power exactly as it is multiply by the natural logarithm of the base and then because in the exponent I have a function of X I then have to multiply by the derivative of that exponent the derivative of 3x + 1 is just 3 if you made it to the end of the video good job hopefully it helped you out and let me know in the comments what you want a top 10 of next Jen MO