in this video we're going to focus on logarithms we're going to talk about how to evaluate logs how to use the change of base formula how to expand condense logs solve equations and also graph logarithmic functions so let's begin let's begin with evaluating logs what is log base 2 of 4 when you see a question like this what you really need to ask yourself 2 to the what power is 4 how many 2's do I have to multiply to get to 4 2 times 2 is 4 we need to multiply to 2 to get to 4 so the answer is 2 2 squared is equal to 4 now what about this one what is log base 2 of 8 how many 2's do I need to multiply to get to 8 I need to multiply 3 2's to get to 8 so 2 to the third power is equal to 8 therefore log base 2 of 8 is 3 try this one log base 3 of 9 3 to the what power is 9 3 squared is 9 so log base 3 of 9 is 2 here's another one what's log base 4 of 16 log base 3 of 27 log base 2 of 32 so 4 raised to the what power is 16 4 squared is 16 4 times 4 16 how many 3's do we have to multiply to get to 27 3 times 3 times 3 is 27 so you need 3 threes to get to 27 3 to the third power is 27 now what about log base 2 of 32 how many twos do you got to multiply to get to 32 2 times 2 is 4 times 2 is 8 times 2 is 16 times 2 is 32 so we need five 2's to get the two 32 2 to the fifth power is 32 now what about log base 5 of 125 log base 6 of 36 log base 2 of 64 and log base 3 of 81 feel free to pause the video and work out these examples so how many fives do we need to multiply to get to 125 5 to the third power is 125 6 raised to the what power is 36 it turns out 6 squared is 36 because the only two multiply to 6 is against 36 how many twos do we need to multiply to get to 64 so least a two to the fifth power is 32 if you multiply 32 by 2 you get 64 so we need 6 2's 2 to the sixth power is 64 now what is log base 3 of 81 it turns out that 3 to the fourth power is 81 if you multiply four threes together that's 81 3 times 3 this is 9 and these threes turn into a 9 as well 9 times 9 is 81 so log base 3 of 81 is equal to 4 since 3 to the fourth power is 81 try these 2 log base 7 49 and log base 3 of 1 we know that 7 squared is 49 so this is equal to 2 now what is log base 3 of 1 lakh of 1 is always 0 regardless of the base because 3 to the 0 power is 1 anything raised to the 0 power will always be 1 now if you see a question on the test that asked you to evaluate log 10 what is the answer whenever you don't see a base it's assumed to be a 10 so 10 raised to the what power is 10 this is 1 log of 10 is 1 now what about log of 100 it turns out log of 100 is 2 because 10 squared 10 times 10 is 100 now what about log of a thousand ten to the what power is a thousand ten times ten times ten three times as a thousand so ten to the third power is a thousand does log of thousands three now what is log of 1 million so here's a simplified way of getting the answer count the zeros here we have 1 0 so the answers 1 here it's 2 0 so 2 2 and here we have 3 zeros so it's 3 in this example we have a total of 6 zeros so log 1 million is equal to 6 a million is basically 10 to the 6 power a thousand is 10 to the third power what about these log of 0.1 log point 0 1 and log of point 0 0 1 a log of 0.1 is equal to negative 1 when it's less than 1 it's going to be negative log of point 0 1 is negative 2 and log of point 0 0 1 is negative 3 simply count the zeros include in the 0 to the left so knowing that what do you think log of point 0 0 0 0 1 is so if we count the zeros 1 2 3 4 5 this is equal to negative 5 by the way what do you think log of 0 and log of negative 5 is equal to negative 5 the log of negative 5 does not exist you cannot have a negative number inside a log and log of 0 doesn't exist so the number inside of log let's say if you have log X X has to be greater than 0 not equal to it but greater than 0 so you can have a point zero-one inside of a log but you can have zero or negative number consider this problem log base 4 of 16 try that one and also try these log base 4 of 1 over 16 log base 16 of 4 and log base 16 1 over 4 so what are the answers to these problems so for race to the what power is 16 we know that 4 times 4 is 16 so the answer is 2 now all of these other answers they contain a 4 and a 16 so there's going to be a 2 somewhere now 4 to the what power is 1 over 16 it turns out it's negative 2 since 4 squared is 16 4 raised to the negative 2 is 1 over 16 whenever you have a negative exponent the fraction will flip so instead of being 16 over 1 it's 1 over 16 now what if you reverse in what's log base 16 of 4 so looking at the first one log base 4 of 16 is 2 over 1 if you reverse the numbers to log base 16 to 4 this is going to be 1/2 now what about log base 16 1 over 4 this is negative 1/2 so here's what need to know if you have a fraction within a log the answer is going to be negative now the second thing you want to know if the number inside the log if it's bigger than the base then you're going to have a whole number typically a number greater than 1 however if the base is larger then you're going to have a fraction a number less than 1 try these examples log base six of 216 six of run over 216 log base 2 16 1 over 6 and log base 2 16 6 so how many sixes do you need to multiply to get to 216 it turns out that 6 to the third power is equal to 216 so therefore this is equal to 3 so every other answer is going to have a 3 in it it could be negative it could be positive it could be a fraction you just have to know which one is which so which ones will be negative the ones that have a fraction within the log it's going to be negative so we can put the negative sign for these two and the other ones will be positive now which one will contain a fraction the ones that are going to have the fraction is when a base is larger than what's inside the log so these two will have a fraction now it's either going to be 3 over 1 or 1 over 3 that's the only possibility so the ones with fractions will be 1 over 3 this one is simply going to be 3 just negative 3 so whenever you have a fraction it's negative and whenever the base is larger it's going to be a fraction the 3 is going to be on the bottom and so that's it so if you know the first one you could figure out the rest now the next thing that we're going to go over is the change of base formula log base a of B is equal to log base B over log a and you could change the base to any number so you can make it a new base base e so how can we apply this so we said that log base 4 of 16 is equal to 2 if you have a calculator with you type this in log 16 over log 4 you'll also get the answer to that's how you can evaluate a log that looks like this with a calculator just type in log 16 over log 4 you can try another example we know that log base 2 of 8 is equal to 3 so if you type in your calculator log 8 divided by log 2 you should also get 3 make sure you enclose these numbers in parentheses like this now let's say if you have the expression log base 5 X minus 1 how can you use the change of base formula to convert this log expression into a natural log so using the change of base formula could say it's log X minus y over log 5 now we could make it any base to convert a log into a natural log u want base E natural logs and logs are the same thing but simply natural logs have the base e logs could have any base 5 7 2010 it could be anything but using the change of base form we can make this 7 if we want we can make it 4 we can make it 3 we can make it our but we're going to choose to make it a log base E is natural log so this is equivalent to natural log of X minus y over natural log of 5 so that's how you can convert a log into a natural log using the change of base formula there are some properties of logs that you need to be familiar with for example log a plus log B is equal to log a times B so you can combine two logs into a single log by multiplying what's inside of logs another rule you need to know is log a minus log B is equal to log a divided by log B and also log a squared is equal to 2 log a so if you have an exponent when you can move it to the front that's another property of logs consider the following expression log X plus log y minus log Z how can we combine this log expression into a single log if there's a positive sign in front the variables will go on top if there's a negative sign in front it's going to go on the bottom so it's going to be log X Y over Z that's all you got to do for this particular drum now let's try another example let's say if we have log X minus log Y plus log Z minus log R how would you condense it into a single log expression so log X and log z are positive so we're going to put X and Z on top log Y and log R contain a negative sign in front so y + R is going to go on the bottom so it's log X Z divided by Y R here's another one - log X + 3 log Y - 4 log Z condense the following expression into a single log so this time we have coefficients right now we can't condense it into a single log because of the coefficients so we need to use the power rule remember we said that log a squared is the same as 2 log a but what we need to do is take the 2 and move it back to the exponent position so the 2 it's going to be the exponent of X the 3 is going to be the exponent of Y and 4 is going to be the exponent of Z so it's log x squared plus log Y cubed minus log Z to the fourth so now at this point we can combine it into a single log expression so x squared + y cube have a positive sign in front of them and z2 the fourth has the negative sign so it's going to go on the bottom so this is it's log x squared Y to the third over Z to the fourth now let's try one more example on condensing log expressions into a single log so try this one just like the last example we need to move the coefficients to their respective exponent positions so this expression is equivalent to log X raised to the one-half minus log raised to the one-third plus log Z raised to the 1/4 so now at this point we can combine it into a single log expression so X and Z will be on top since they have a positive sign in front so it's X to the one-half Z to the one-fourth now the Y variable is associated the negative sign so it's going to be on the bottom now because we have fractional exponents we can convert it to a radical expression X to the one-half is the same as the square root of x z to the one-fourth is equivalent to the fourth root of z to the first power or simply just a z and the cube root of y we can write it this way so this is the answer in radical form how can we expand the following logarithmic expression so this time we're working backwards we want to expand the log into multiple logs so log R squared s to the fifth Z to the sixth power is the same as log R squared plus log S to the fifth power minus log Z to the six we have a negative sign since C is on the bottom now what we need to do is move each exponent back to the front so the final answer is two log R plus five log s minus six log Z try this one log cube root x squared Y over Z to the fourth so how can we expand this particular expression the first thing I would do is rewrite it as log x squared Y Z to the fourth raise the 1/3 that's the same as the cube root of what's inside so before we can expand it we need to move the 1/3 to the front so it's going to be one-third times everything else so now we can expand the logarithmic expression now the 1/3 is going to be distributed to each of the three new logs so it's going to be log x squared plus log Y minus four log Z since C is on the bottom so now we'll leave it as log Z to the fourth the last thing you need to do is take the exponents and move it to the front so the final answer is going to look like this is one third to log X plus log Y minus four log Z this is it that's how you can expand the logarithmic expression but let's try another one try this practice from the first thing I will do is convert the radical expressions into exponents the square root of Y is y to the 1/2 the cube root of Z to the fourth is Z to the 4/3 so now we can expand the expression so it's log x squared plus log Y to the 1/2 minus log Z to the 4/3 so now let's move the exponents back to the front so the final answer is 2 log X plus 1/2 log Y minus 4/3 log Z and that's it for this problem now let's simplify a few expressions what is the value of L M 1 Ln E and Ln e to the fifth power the natural log of 1 or the log of 1 is always it's always zero the natural log of e is always equal to 1 now what do you think Ln E to the fifth power is equal to notice that the 5 we can move it to the front so Ln E to the fifth power is 5 ln e and keep in mind ln e is 1 so this is simply equal to 5 try these e to the ln 7 e to the 3 ln x + 2 e 4 ln 1 so what's e to the ln 7 notice that the base of natural log is e whenever in a base is raised to an exponent that contains the base of itself those two bases will cancel and the answer is simply 7 for this one we need to take the three and move it to the exponent of X e to the three Ln X is the same as e to the Ln X cubed and so this is base e these two will cancel and is simply X cubed now what about this one this is the same as two e to the Ln Y to the fourth now this part right here will change to Y to the fourth so the answer is simply two times y to the fourth try these two problems so what is five raised to the log base 5 of Y to the 8th so just like the last examples these will cancel and the answer is simply y to the 8th power for this one the 3s will cancel and it's x squared plus 6 so you now know what to do if you ever see something like that on the test simplify the following log expression how would you simplify what's log base 3 of 6 minus log base 3 of 2 whenever you wish to subtract two logs you can combine them into a single log by means of division so it's equivalent to log base 3 6 divided by 2 6 divided by 2 is 3 now what is log base 3 of 3 3 to the what power is equal to 3 this is 3 to the first power 3 to the first power string so log base 3 of 3 is 1 keep in mind you could change it to you can use the change of base formula log base 3 of 3 is the same as log 3 over log 3 which cancels and give you 1 I'll try another example try this one log base 2 of 12 plus log base 2 of 24 minus log base 2 of 9 so feel free to pause the video and see if you can get the answer so let's combine it into a single log so the 12 and the 24 are associated for positive sign so it's going to be 12 times 24 on top the 9 is associate of a negative sign so we're going to put a 9 on the bottom so what we need to do is simplify this expression so what's 12 times 24 divided by 9 let's break it into smaller numbers 12 is 4 times 3 24 is 8 times 3 and 9 is 3 times 3 so we can cancel the threes and 4 times 8 is 32 so what we really have is log base 2 of 32 now how many twos do we need to multiply to get to 32 keep in mind two to the 5th power is 32 so log base 2 of 32 is 5 here's another one log base 7 1 over 6 plus log base 7 6 over 49 so what can we do in this particular problem how can we solve it so just like before since we have two logs separated by an addition sign we're going to combine it into a single log by multiplying what's on the inside so it's 1 over 6 times 6 over 49 notice that the 6 is they cancel and so what we have left over is log base 7 over 1 of 1 over 49 so we know that 7 squared is 49 so therefore 7 to the negative 2 power is 1 over 49 therefore this expression is equal to negative 2 so that's the answer sometimes you need to change an expression from logarithmic form into exponential form how can we do that 2 to the third power is equal to 8 and that's how you can do it try this one 3 to the second power is equal to 9 five squared is equal to 25 so that's a simple way in which you can convert a log expression into exponential form try these two a log base a of B is equal to C and log base two of X is equal to Y so a race to the C is equal to B that's how you could change it to exponential form 2 raised to the Y is equal to X so now let's convert back to logarithmic form convert each exponential equation into a log so what this really means is log base 3 of 27 is equal to 3 if you work backwards 3 to the third is 27 so the exponent always goes across the equal sign the base that has the exponent is the base of the log so this expression the next one 2 to the fourth is equal to 16 is the same as log base 2 of 16 is equal to 4 so the exponent is what the log is equal to and this is the base of the exponent which is the base of log so here we have log base 3 of 18 is equal to X so X is the exponent and the base is 3 now let's spend some time solving logarithmic equations log base X of 27 is equal to 3 solve for X so let's convert it to exponential form X raised to the third power is 27 so clearly we can see X is 3 because 3 to the third is 27 therefore X is equal to 3 try this one log base 2 of Y is equal to 5 so if we convert it to exponential form 2 raised to the fifth power is equal to Y 2 to the fifth power is 2 times 2 times 2 times 2 times 2 five times two and two makes 4 and 4 times 4 is 16 16 times 2 is 32 so Y is equal to 32 try this problem go ahead and solve for X feel free to pause the video as you work through these examples two to the third power is equal to what's inside so two to the third power is equal to X plus five two times two times two is eight so to solve for X all we need to do is subtract both sides by 5 and 8 minus 5 is 3 therefore X is equal to 3 let's try another problem like that log base 3 of 2x minus 3 is equal to 2 so 3 raised to the second power equals what's inside so 3 squared is equal to 2x minus 3 3 times 3 is 9 so now we just need to solve for X let's add 3 to both sides of the equation 3 plus 9 is 12 since 2 is multiplied to the X to separate the X from the 2 we need to perform the opposite of multiplication which is division so we need to divide both sides by 2 12 divided by 2 is 6 and therefore X is equal to 6 try that one so what do you think is the first thing we need to do to solve for X what we should do is to condense the two logs into a single log by means of multiplication so it's log base 2 x or log base 2 of X minus 3 times X minus 1 is equal to 3 so now we can convert it to exponential form 2 to the third power equals what's inside so 2 to the third power is equal to X minus 3 times X minus 1 on the right side and need to foil X times X is x squared and x times negative 1 that's a negative X negative 3 times X is negative 3x and negative 3 times negative 1 that's plus 3 2 to the third power is 8 so now let's combine like terms on the right side negative 1x and negative 3x when added they will change to negative 4x so now we need to make some space I might need the original expression so let's subtract both sides by 8 if we do that we will get a 0 on the left side and on the right side 3 minus 8 is negative 5 so we need to factor the trinomial which has the leading coefficient of 1 so you need to look for two numbers that multiply to negative 5 let that add to negative 4 what are those two numbers the two numbers are negative 5 & 1 negative 5 times positive 1 is negative 5 and negative 5 plus 1 is negative 4 so we can simply factor it like this it's going to be X minus 5 times X plus 1 so now we could solve for X let's set each factor equal to 0 so X minus 5 is equal zero and X plus one is equal to zero so for this equation we need to subtract one from both sides but here we need to add five so X is equal to five and X is equal to negative one so now we're dealing with logs you need to find out if you have any extraneous solutions so our both of these answers acceptable or do we have an extraneous solution so keep in mind you can never have a negative log I mean a negative number inside a log so if we plug in five into X minus three five minus three is positive two that's okay if we plug in 5 here 5 minus 1 is 4 that's okay but if we plug in negative 1 negative 1 minus 3 that's negative for log base 2 of negative 4 it doesn't exist so negative 1 is not in the domain of the log function so therefore X can't equal negative 1 this is the extraneous solution and this is the actual solution the one that we want so the answer is X is equal to 5 sometimes you might get a problem that wants you to identify the extraneous solution which is this one try this one log base 3 of 2x plus 1 plus log base 3 X plus 8 now let's say that's equal to 3 feel free to pause the video and work this problem so let's combine the two log expressions into a single log so we need to multiply 2x plus 1 and X plus 8 so now let's change it to exponential form 3 to the third power is equal to what's inside so 3 to the third is equal to 2x plus 1 times X plus 8 3 to the third that's 3 times 3 times 3 that's 27 and that one into foil 2x times X is 2x squared 2x times 8 is 16 X 1 times X that's X and 1 times 8 is 8 so now let's add like terms 16x plus X that's equal to 17 X so let's subtract both sides by 27 so zero is equal to 2x squared plus 17 X 8 minus 27 that's about negative 19 so let's see if we can factor this particular expression how would you factor 2x squared + 17 X minus 19 so notice that we have a trinomial where the leading coefficient is not 1 so we have to take into account the to multiply the leading coefficient by the constant term that's 2 times negative 19 and that gives you negative 38 what two numbers multiply to negative 38 but add to positive 17 and that's going to be positive 19 and negative 2 19 times negative 2 is negative 38 by 19 plus negative 2 is positive 17 so what we're going to do is we're going to replace the middle term with negative 2 X + 19 X now at this point what you want to do is you want to factor by grouping you want to factor out the GCF the greatest common factor from the first two terms so what can we take out from - x squared + 2 X the most that we can take out as a 2 X - x squared divided by 2 X is equal to X and negative 2 X divided by 2 X is equal to negative 1 now from 19 X and negative 19 we can take out a 1919 X divided by 19 is equal to X negative 19 the vitamine I mean is equal to negative 1 notice that we have a common factor X - 1 when you see that that you're on the right track so let's factor out X minus one now what's going to go inside the settling parenthesis it's going to be what's outside of X minus 1 the 2x and the plus 19 so we can put that here so now we can solve for X so let's set each factor equal to 0 and let's solve here we need to add 1 to both sides so our first answer is X is equal to 1 for the next one we need to take away 19 from both sides so 2x is equal to negative 19 and then we need to divide both sides by 2 so the other answer is X is equal to negative 19 over 2 so we can clearly see that we can plug in 1 into 2x plus 1 and X plus 8 and that's not going to give us any negative answers however if we plug in negative 19 / - that's like negative nine point five negative nine point five plus eight is negative nine point one so this is an extraneous solution so we can get rid of it so the answer is one now let's check the work let's see if we have the right answer let's see if it's correct so let's plug in 1 into the equation so it's going to be two times one plus one and on this side one plus 8 so 2 times 1 is 2 plus 1 that's 3 so we have log base 3 of 3 1 plus 8 is 9 now log base 3 of 3 the 3s cancel so it's simply 1 what's log base 3 of 9 3 squared is equal to 9 so this is equal to 2 & 1 plus 2 is 3 so which means that X is indeed equal to 1 both sides of the equation have the same value so here's another one log base 2 x + 6 - log base 2 X - 8 is equal to 3 go ahead and solve for X so because we have a minus sign it's going to be division so when we combine it into a single log it's going to be X plus 6 on top and X minus E on the bottom and this is equal to 3 so 2 raised to the third power is equal to what's inside 2 to the third power is the same as 8 and we can write it as 8 over 1 so 8 over 1 is equal to X plus 6 divided by X minus 8 so let's cross multiply eight X minus 64 that's 8 times X minus 8 and then 1 times X plus 6 is simply X plus 6 on the other side so let's subtract both sides by X and simultaneously let's add 64 to both sides so the X variables will cancel and that 2 8 X minus 1 X is 7 X 6 plus 64 is 70 so if we divide both sides by 7 70 divided by 7 is 10 therefore X is equal to 10 consider this example log base 2 of X plus 5 is equal to log base 2 of 3x minus 9 how can we solve for X in this particular example notice that we have the same base same log therefore what's inside should equal each other it turns out that X plus 5 we could set it equal to 3x minus 9 whenever you have a log equivalent to another log on the other side of the equation you could set the inside equal to each other if they have the same base so let's subtract both sides by X and at the same time let's add 9 to both sides so 5 plus 9 is 14 3x minus 2x I mean 3x minus 1x is 2x so 14 divided by 2 at 7 therefore X is equal to 7 how would you solve for X let's say Ln X is equal to five what is the value of x so what you need to know is the base of the natural log the base of the natural log is e and if we convert it to exponential form earase to the fifth power is equal to x and so we have the value of x is e raised to the fifth power try this one Ln X plus three Ln X is equal to zero so if you don't see a coefficient it's a 1 1 plus 3 is 4 so 4 Ln X is equal to 0 if we divide both sides by 4 4 divided by 4 is 1 so we simply have Ln X on the left side a 0 divided by 4 is 0 so e raised to the 0 is equal to X anything raised to the 0 power is 1 so X is equal to 1 keep my natural log of 1 is 0 here's another example you can work on if the natural log of X minus 3 is equal to 2 what is the value of X so for this example let's convert it to exponential form a squared is equal to X minus 3 so to solve for X we simply need to add 3 to both sides so X is equal to e squared plus 3 now sometimes you might be given exponential equations to solve how would you solve for X in this particular equation in what you want to do is see if you can convert the bases into a common base the common base of 8 and 16 is 2 2 to the third power is 8 2 to the 4th power 16 so we can replace 8 with 2 to the third and 16 with 2 to the fourth now what happens when you raise 1x1 into another whenever you raise an exponent to another exponent you should multiply them so 3 times X plus 4 is 3x plus 12 4 times 2x minus 3 is 8 X minus 12 so now we could solve for X since the bases are the same the exponents must be the same therefore 3x plus 12 must equal to 8x minus 12 so we could solve for x let's add 12 to both sides and let's take away 3 X from both sides 12 plus 12 is 24 8 X minus 3x is 5x so therefore X is equal to 24 over 5 and now it's your turn try this example so what is the common base of 9 and 27 it turns out that 3 squared is 9 and 3 to the third power is 27 so let's replace 9 with 3 squared and 27 with 3 to the third now let's multiply the exponents so 2 times 2x minus 1 that's going to be 4 X minus 2 and 3 times 3 X plus 6 that's a 9 X plus 18 if we distribute the 3 so now we can set the exponents equal to each other so 4 X minus 2 is equal to 9 X plus 18 so let's add 2 to both sides and let's take away 9 X from both sides so 4x minus 9 X is negative 5 X 18 plus 2 is 20 so if we divide both sides by negative 5 X is equal to 94 and so this is the answer how would you solve for X what is the answer we know that 2 squared is 4 and 2 to the 3rd power is 8 so X has to be somewhere between 2 & 3 because 7 is between 4 & 8 so X is like 2 point something but how can we find the value of x we can't change 7 into a base 2 so in this particular instance to solve for X you want to use logs you can use a log or natural log you'll get the same answer so I'm going to take the natural log of both sides once I do that I can take the X variable and move it to the front so X Ln 2 is equal to Ln 7 so to solve for X divide both sides by Ln 2 so now we have the answer X is equal to the natural log of 7 divided by the natural log of 2 so this point you need a calculator Ln 7 divided by Ln 2 you should get about 2 point 8 0 7 3 5 now you can check your work if you type in 2 raised to the 2 point 8 0 7 3 5 the answer should be close to 7 once I type it in I get 6 point 9 9 9 9 7 6 which is approximately 7 so this technique works so that's how you can solve for X when it's an exponent you can use logs and natural logs try this one go ahead and solve for X so if the base is e you should definitely use natural log as opposed to the regular logs so let's take the natural log of both sides and the reason why you want to use the natural log is because Ln E is 1 now let's move this to the front so we have X plus 2 times ln e is equal to ln 8 when we said ln e is 1 so this disappears so we simply have X plus 2 on the left side X plus 2 times 1 is X plus 2 so all we got to do is subtract both sides by 2 and the answer is X is equal to ln 8 minus 2 consider the function f of X is equal to natural log 2x minus 6 if you're given this function how can you find the domain of this function now keep in mind y is the same as f of X so you can view this equation as y is equal to natural log 2x minus 6 so what is the domain now you need to remember that you can't have a negative number or a zero inside a log or natural log therefore 2x minus 6 must be greater than zero so you could solve for x so if we add 6 to both sides 2x is greater than 6 and then if we divide by 2 we can see that X is greater than 3 so the domain is from 3 to infinity and that's all you need to do if you only have one critical point in this case 3 is the critical point let's say if you have two critical numbers it's not a symbol for example go ahead and find the domain for this function the natural log of x squared plus 2x minus 15 feel free to pause the video take a minute and see if you can find the domain for that function so the first thing I would do is set the inside greater than zero so let's solve for X we need to factor what two numbers multiply to negative 15 but adds in a middle coefficient to this is going to be negative 3 and positive 5 negative 3 times 5 is negative 15 but negative 3 plus 5 is positive 2 since the leading coefficient is 1 we can simply write it as X minus 3 times X plus 5 so there's two critical numbers 3 for X and negative 5 for X as well but how do you use those numbers to determine where the function is going to be greater than 0 or less endure particularly the inside part of the function since we have two critical numbers what we need is a number line we need to create a sign chart we got to test the numbers to see what works and what doesn't so the first critical number is negative 5 and the second is positive 3 don't forget to switch the signs if you solve for each one if you set X minus 3 equal to 0 X will be plus 3 if you set X plus 5 equal to 0 and if you solve for X X is negative 5 so you have to change the signs all the way to the left we have negative infinity and all the way to the right we have infinity so let's plug in some test points if we plug in a number between 3 to infinity let's say if we plug in 4 into this expression is it going to be positive or negative so 4 minus 3 is a positive number 4 plus 5 is a positive number whenever you multiply two positive numbers you will get a positive result so let's put a plus for this region now pick a number between negative five and three let's use zero so zero minus three is negative zero plus five is a positive number whenever you multiply a negative number and a positive number you're going to get a negative number now if we plug in a number between negative five and negative infinity let's say negative six negative six minus three is negative negative six plus five is negative two negative numbers when multiplied will give you a positive number so keep in mind we want the portion of this equation that is greater than zero that means the positive part and not the negative part so we want this region and this region that's where the equation works anywhere between negative five and three the function will have a negative value and you can have a negative number inside a natural log so it's not going to work in this region so therefore the domain you can literally see it it's going to include this region and this region so the domain is from negative infinity to negative five Union 3 to infinity now does it include five or should we exclude it notice that it's greater than zero not greater than or equal to zero so we have an open circle at negative five and three and therefore we should use parentheses and not brackets at negative five and three so that's how you can find the domain of a natural log function or even a log function so if you have one critical number simply set it greater than zero and solve for x you're done if you have two more critical numbers you need a number line or a sign chart to figure this up consider the function Ln X plus three minus one given this function how can you find the inverse function what do you need to do what steps do you have to take so the first thing you want to do is replace f of X with Y f of X and y are the same thing your next step to find the inverse function you want to switch x and y so X is equal to Ln y plus three minus one after that you want to solve for y so let's add one to both sides so X plus one is equal to Ln y plus three to get rid of the natural log we need to convert it to exponential form so e raised to the X plus one is equal to what's inside so therefore we have this equation e to the X plus one is equal to y plus three so now the last thing we need to do is subtract three from both sides so Y is equal to e raised to the X plus one minus three at this point you can replace this Y with the inverse function symbol so f to the negative one x which is the inverse function is e to the X plus one minus three and so this is the answer by the way notice that natural log functions and exponential functions are inversely related as you can see we started with a natural log function and the inverse function was an exponential function so what we're going to do at this point is we're going to graph a few exponential functions and then a few logarithmic functions so let's start with y is equal to 2 raised to the X minus 3 plus 1 so what do we need to do to graph the function let's take the exponent and let's set it equal to two numbers zero and one and solve for X if X minus three is equal to zero X is equal to three and if X minus three is equal to 1 X is 4 now those points I'm going to choose for my table I'm going to choose 3 & 4 so what is the value of y when X is 3 so 2 raised to the 3 minus 3 plus 1 3 minus 3 is 0 anything raised to the 0 power is 1 so this is equal to 2 so in X is 3 y is 2 now what about when X is 4 what is the value of y so 2 raised to the 4 minus 3 plus 1 4 minus 3 is 1 2 to the first power is 2 and 2 plus one and string so we have the point 4 comma 3 this number that you see on the outside the plus 1 and that's the horizontal asymptote for the exponential function exponential functions have horizontal asymptotes and logarithmic functions have vertical asymptotes so the equation for the horizontal asymptote is y is equal to 1 once you have the asymptotes and two points you have all that you need to graph the exponential function so let's plot the horizontal asymptote at y equals 1 and then we have the point 3 comma 2 which is over here and 4 3 which is around this area so the graph starts from the horizontal asymptote and then it follows the two points the exponential function is an increase in function that goes up at an increasing rate the logarithmic function is an increasing function that increases at a decreasing rate let's try another exponential function go ahead and graph this one so just like before we're going to set the exponent equal to zero and one so if X plus one is zero that means X is negative one and if X plus one is equal to 1 X is zero so we need to plug in 0 and 1 into the equation and let's find the values of Y so e to the negative 1 plus 1 minus 2 negative 1 plus 1 is 0 anything raised to the 0 power is 1 so when 1 minus 2 is negative 1 now if we plug in 0 for X 0 plus 1 is 1 and so we simply have a minus 2 so keep in mind e is approximately about 2.7 so 2.7 minus 2 is about point 7 lastly we need the horizontal asymptote which is basically this number next to e so it's y is equal to negative 2 so this is where the horizontal asymptote is negative 2 and we have the points negative 1 negative 1 and 0.7 which is around here so the graph is going to start from the x-axis and then it's going to increase in that direction so now let's say if you have a graph that looks like this it's 3 minus 2 to the 4 minus X try this one so let's set 4 minus X equal to 0 and 1 so if we add X to both sides it's supposed to be a 1/4 is equal to X and for the other one we have 4 is equal to 1 plus X if we add X to both sides so here we got to take away 4 so X is equal to 4 and also X is equal to 3 so our table should have 2 points 3 & 4 4 minus 3 is 1 & 4 minus 4 is 0 so now let's plug in the points into the equation so 3 minus 2 times 4 minus let's start with 3 so 4 minus 3 is 1 3 minus 2 is also 1 so when X is 3 y is 1 now if we plug in 4 it's going to be 4 minus 4 which is 0 2 to the 0 power is 1 3 minus 1 is 2 when you do it this way you will always get numbers that you can work with now what is the horizontal asymptote for this graph the horizontal asymptote is the number outside of the exponential part of the function so it's positive 3 y is equal to positive 3 don't let this negative sign confuse you that sign is for the 2 in front of the 3 it's positive so that's why it's positive 3 so now let's graph it so let's start with the horizontal asymptote at y equals three and then we have the point 3 1 which is over here and also for 2 now keep in mind the graph always starts from the horizontal asymptote this time it's going to start here and go this direction so notice that we have a negative sign in front of the X when you see that it reflects over the y-axis and also we have a negative sign in front of the 2 which reflects over the x-axis so let me give you all the different shapes and their transformations so if you have Y is equal to a positive 2 raised to the positive x so when x and y are positive let's say this is for X and this is for y x and y are positive in quadrant two I mean that quadrant excuse me quadrant 1 so the graph is going to go towards quadrant 1 now when Y is positive Bolen x is negative it's going to go towards quadrant 2 so it goes this one in quadrant 2 x is negative you got to go to the left but Y is positive because you're going up now what quadrant is it going to go to if Y is negative but X is positive in this case it's going to go towards quadrant 4 because X is positive towards quadrant 4 because you're going to the right and Y is negative because you're going down now what about when X and y are both negative then it's going to go towards quadrant 3 which is the example that we have in Quadrant 3 you go to the left because X is negative and you're going down so Y is negative so it's a reflection about the origin so to speak whenever you reflect over the x-axis and the y-axis is the same or equivalent as you flex it over the origin so let's move on to the graphs of logarithms how can we graph it so for logs set the inside part the X minus 3 equal to 3 things 0 1 and whatever the basis in this case the base is 2 so we're going to set it equal to 2 now when you set it equal to 0 it's not going to give you a point rather this for the vertical asymptote so if you add 3 to both sides you can see that the vertical asymptote is at x equals 3 now the other two points is for the table so if X minus 3 is 1 if we add both sides if we have 3 to both sides one plus three is four so we get the point four for X and if X minus 3 is 2 if we add 3 to both sides 2 plus 3 is 5 so X is 5 so now let's plug in these points 4 minus 3 is 1 log of 1 is always 0 and 0 plus 1 is 1 now let's replace X with 5 5 minus 3 is 2 the log base 2 of 2 because the numbers are the same they're going to cancel it's simply 1 so 1 plus 1 is 2 when you do it this way typically your Y values will be integers so now we have enough information to graph the function so the vertical asymptote is that X is equal to 3 and we have the point for one and we have the 0.52 so this time instead of starting from a horizontal asymptote the graph is going to start from the vertical asymptote and it's going to follow those two points so it's going to go like this and so that's how you can graph a logarithmic expression now what is the domain for this function notice that the lowest x value is 3 and the highest is infinity so the domain is from 3 to infinity now what about the range the range for a logarithmic function or even a natural log it's always negative infinity to infinity it's always going to be the case now let's briefly go back to the exponential function that we had in the beginning so we said that it had a horizontal asymptote at y equals 1 and it was an increasing function I'm going to draw a rough sketch it was something like this what is the domain and range for this function I forgot to mention earlier so the domain for an exponential function is always going to be arvo numbers the range for a logarithmic function is all real numbers the range for an exponential function is limited it's limited by the horizontal asymptote the lowest Y value is 1 and the highest goes all the way to infinity but it doesn't touch 1 so it's 1 to infinity but does include 1 so now there was another exponential equation that looked like this so we had a horizontal asymptote at 3 and because x and y are negative this graph is going to start from the horizontal asymptote and go towards quadrant 3 so what is the domain and range for this function because it's an exponential function the domain is all real numbers the range however is different the lowest Y value is negative infinity because the function is going down the highest is the horizontal asymptote 3 so the range is from negative infinity to 3 not including 3 try this one natural log X minus 1 plus 2 go ahead and graph the equation so we're going to set the inside equal to 0 1 and the base the base of a natural log is e now let's make the table so for the first one is for the vertical asymptote that's the one where you set it equal to 0 so if X minus 1 is equal to 0 that means X is equal to 1 now for the second 2 if X minus 1 is equal to 1 if you add 1 to both sides 1 plus 1 is 2 X is 2 and for the other one it's going to be 1 plus e so now let's find the Y values that correspond to those points so let's start with 2 natural log of 2 minus 1 plus 2 2 minus 1 is 2 I mean excuse me 2 minus 1 is 1 not 2 log of 1 is always 0 so 0 plus 2 we have the point 2 comma 2 now let's plug in 1 plus e so 1 plus e minus 1 plus 2 so the ones will cancel and we're simply going to have Ln e Ln E is always equal to 1 1 plus 2 is 3 so we have the point 1 plus e comma 3 now keep in mind a is 2.7 so 1 plus 2.7 the x value is about 3.7 so now let's graph it so we have a horizontal asymptote at y equals two actually not a horizontal asymptote I take that back you only have horizontal asymptotes for exponential functions I'm looking at this number but we have a vertical asymptote at one logarithmic functions always have vertical asymptotes and then we have the point 2 comma 2 which is over here and then three point seven three which is roughly about there so the graph looks something like this so what is the domain and range the range for a logarithmic function is always negative infinity to infinity as you can see the lowest Y value is negative infinity and the highest is positive infinity now for the domain is restricted the lowest x value is the vertical asymptote so it starts that one and the highest x value is positive infinity so the domain is from one to infinity try this log base 3 2 minus X plus 1 so what I want you to do for this example graph the equation find the asymptotes and then also graph the inverse function so let's begin let's set the inside of the log equal to 0 1 and the value of the base in this case the basis stream so let's make our table the vertical asymptote is for the first one so if 2 minus x is 0 X has to be 2 because 2 minus 2 is 0 if 2 minus X is 1 X is 1 2 minus 1 is 1 now what about 1 2 minus X is 3 what is the value of X so if we subtract both sides by 2 negative x is 1 which means X is negative 1 so now let's find the Y values that correspond to those points 2 minus 1 is 1 log of 1 is always 0 so 0 plus 1 is 1 now let's plug in negative 1 2 minus negative 1 that's the same as 2 plus 1 which is 3 a log base 3 of 3 the 3s cancel you get 1 1 plus 1 is 2 so now let's graph it so the vertical asymptote is at two and we have the points 1 1 which is over here and negative 1 2 so the graph is going to start from the vertical asymptote and it's going to follow these two points which I need to do that one more time so it's going to look like this notice that we have a negative in front of the X and a positive in front of the log so negative x positive Y that's quadrant two so it's going to go to the left and up now let's go ahead and graph the inverse function the vertical line for the original function is X is equal to 2 for the inverse we need to switch X with Y so the horizontal asymptote it's going to be y is equal to 2 for the inverse function so instead of X is equal to 2 is now Y is equal to 2 now notice that the points that we have is 1 1 so if you switch x and y we're still going to get a point at 1 1 but I need to use a different color so I'm going to use yellow for the inverse function now for the second point negative 1 2 and we need to change it to 2 negative 1 so X is 2 y is negative 1 so 2 negative 1 should be somewhere in this region so to plot the exponential function we need to start from the asymptotes and it's going to follow these two points it's not bounded by the vertical asymptote notice that those two graphs the one in green and yellow they're symmetrical about the line y equals x which is true of any inverse function now if we had a typical log let's say this is the vertical asymptote of x equals two and let's say the function went this way instead for the exponential function which is the inverse of a log function it's going to go like this so you could see the symmetry across this line so this is y is equal to log probably like X minus 2 and for this one it's e to the X plus 2 and so those two functions are inverses of each other they're symmetrical about the line y equals x they reflect across that line so that is it for this video thanks for watching and