by the end of this video you'll easily be able to chop up these log equations and in about 30 seconds here's how this video is going to get you there we're going to start off with a pretty quick log equation but these are quickly going to get harder and harder as we go we're going to start seeing some new things like what happens when we get a natural log in our equation and then problems five and six especially are gonna have a lot more work and after we go through all of that i'm gonna give you a problem to try and answer in the comments and by that point you should be able to chop this log up into some firewood and make some s'mores with it and if you're looking for the notes for this video and when i say that i mean printable notes i have printable notes for this video and if you're looking for those they're linked right in the description also in the description i have an extra video where you and i will go through and solve 12 more log equations so if you're looking for some more practice especially if you have a quiz or a test coming up on this kind of stuff then i highly recommend you check out that extra video in the description so starting with this first problem we have the log of 4x minus 5 is equal to the log of 2x minus 1. so how are we going to solve a log equation like this well this is honestly the nicest kind of log equation because the only way for these two logs to be equal is if their arguments are equal if 4x minus 5 equals 2x minus 1. and you'll notice i use the word arguments there the arguments of the logs are the things that you're taking the log of so 4x minus 5 and 2x minus 1 here now we know that these things have to be equal because if you have the log of 2 this isn't going to be equal to like the log of 3 or the log of 4 or anything like that no they're definitely not equal the only log that the log of 2 is going to be equal to is the log of 2. so these numbers have to be the same for these two to actually be equal so that's how we know that the arguments have to be equal and so what we can do from here is just solve for x so let's start by subtracting 2x on both sides that'll get all my x's to the left-hand side and then if i'm getting all my x's to the left-hand side i want to get all my non-x's to the right-hand side so this negative 5's got to go and so we can add it on both sides so on the left we get 4x minus 2x that's 2x on the right we get negative 1 plus 5 that's 4 and then dividing by 2 on both sides you can see that x is equal to 2. so that's our first log equation already done now moving on to problem 2 we have negative 10 plus log base 3 of x plus 3 is equal to negative 10. and you can see that this problem looks a little bit different than problem 1. we only have one of our terms here that is actually a log other terms like this negative 10 and this negative 10 they don't have logs on them and so this problem is going to solve differently than something like problem one where everything that we were dealing with was a log and these problem one and problem two these make up the two main kinds of problem types that you can see with log equations you can either have everything be a log in which case what your goal is going to be is to get a log on one side and a log on the other side or not all of your terms will be logs only one of them was a log in this case and when that happens what you want to do is you want to isolate your log on one side and get everything else that's not a log to the other side and so here that means we need to add 10 on both sides to get that 10 to the other side and doing that we're going to get a log base 3 of x plus 3 is equal to negative 10 plus 10 which is 0. so we've gotten that negative 10 to the other side now and so our log is isolated and from here we can just evaluate it so we're going to make this 3 bigger and we put it here that zero comes over and it is now the power on three and the x plus three goes to the other side and if you don't feel 100 comfortable with that whole evaluating logs process then i did make a whole youtube video on that but continuing on here anything to the zero power is one so three to the zero is one and so we get one is equal to x plus three and subtracting three on both sides we get that x is equal to negative two and that is the answer for problem two moving on to problem three now which kind of problem is this more like is it's more like problem one or is this like problem two more well looking at all of our terms here they are all logs and so what that means remember is we want to get a log on one side and a log on the other side but we don't want to have two logs on one side and two logs on the other so we want to combine these logs and so how can we do that well looking at the left hand side these two logs are being added and so what that means is we can actually combine them into one log we just have to multiply the arguments and so we get log base 2 of x 2 times x plus 1. now it's the same story on the other side again we have two logs that are being added and so the same rules apply we combine them into one log and just multiply the arguments so now we actually have one log on one side and one log on the other and that means that their arguments must be equal so we know that x plus 2 times x plus 1 is equal to x times 2x plus 4. and now it's a bit more straightforward how you solve for x we can multiply x times x to get x squared x times one gives us plus x two times x gives us a plus two x and two times one gives us plus two and then on the other side we have x times 2x that's 2x squared and x times 4 which is 4x it's from here you can see that we're going to deal with a quadratic we have some x squared terms and so what i want to do is get everything to one side and i want to have a zero on the other and the reason why i drew that arrow to the right is because i can see that if i subtract x squared on both sides and get that x squared over to the right hand side then i'll be left with a positive x squared and i like dealing with that right so i'd ideally like to get a 0 on the left hand side and a positive x squared on the other so x plus 2x that is a 3x and so i can subtract 3x on both sides to cancel off all this stuff that's going to give me 4x minus 3x which is a positive x and then i just subtract 2 on both sides to get my final quadratic here which is x squared plus x minus 2 and this is something that we can factor quickly and solve for x so what are two numbers that add to be 1 and multiply to be negative 2. i'll give you a second to think about well really you only have the options of 2 and 1 here so you just got to figure out where to put the negative sign if you do 2 and negative 1 you found your two numbers so we can write this quadratic as x plus 2 times x minus 1. and from there you can see that your solutions are x equals negative 2 and x equals positive 1. so okay that should be our answers right well not quite there is still something that you need to check for at the end here and actually what we're going to check for is going to eliminate one of these solutions remember that in a log you can't have 0 and you can't have a negative number so you can't have something like the log of 0 and you can't have the log of anything like negative 1 negative 2 negative 3 or any negative number so what that means is that when we plug these solutions back into our log equation we better not get log of 0 or log of a negative number anywhere so let's look let's first plug in negative 2 for x and doing that right away in the first argument here x plus 2 if you plug in negative 2 you get negative 2 plus 2 and that's 0. so right away you can see that's not going to work so what that means is that x equals negative 2 it's a false solution and we call that an extraneous solution but let's look at x equals 1 now can we plug in x equals 1 and get positive arguments all around well if we plug in a one for x in x plus two you get a positive number you get three you'll get a positive number with x plus one you get a positive number with x because you just get one there and then two times one plus 4 is a positive number so that works and x equals 1 is a valid solution and in this case it's our only solution so that's your answer for problem 3. now moving on to problem four what kind of log equation is this well you can see that only one of our terms here has a log we have other terms in this equation that don't have logs and so since not all of our terms are logs what's our goal here well what we want to do is get our log isolated on one side and everything else to the other side so let's start doing that to start isolating this natural log we can subtract 3 on both sides and then i'll get that 3 over to the right hand side and that will leave us with 4 natural log of 2x minus 1 is equal to 11 minus 3 which is 8. now the only thing left to do to fully isolate that natural log is to get that 4 out of there and since it's multiplying the natural log we divide it on both sides doing that we're going to get the natural log of 2x minus 1 is equal to 8 divided by 4 which is 2. and now since we have that log isolated we can do that evaluation step and how do we do that with a natural log well remember a natural log is the same thing as log base e so any time you see ln you can replace it with log base e and that will hopefully clear some things up for you as to what you're actually doing so from here what we can do is make the e bigger drag that 2 over that becomes the power on e and then the 2x minus 1 goes to the other side and from here it's pretty quick to solve for x first we add 1 on both sides and that will give us that 2x is equal to e squared plus 1 and then all we got to do is divide by 2 on both sides and that gives us x is equal to e squared plus 1 over 2 which you can just quickly plug into your calculator if your teacher wants that in the form of a decimal awesome so that is problem four done all right moving on to problem five what kind of problem is this one well you can see that yeah we do have most of our terms are logs but the three is not and so again what is our goal going to be our goal is going to be to get one log on one side and everything else on the other side and so you might be able to see the issue that we have here we have two logs on the left hand side we just want one log and so we need to combine these logs somehow and well these two logs are being added so we can combine them into one log that's a property that we can use the only thing we have to do is just make sure that we multiply the arguments so it's going to be x plus 1 times x minus 1 inside this log base 2 and that's going to equal 3. now from here what we could do is just quickly multiply the x plus 1 through to the x minus 1. doing that you're going to get an x squared x times negative 1 is negative x 1 times x is x and then 1 times negative 1 is negative 1. and you can see here that negative x plus x is 0 so these two go away and you just have x squared minus 1 left and if you really think about it real quick it should make sense to you that what we got is x squared minus one because you know aren't these the factors for x squared minus one if you do difference of perfect squares and so if you recognize that right away you could have went a little bit faster so yeah anyways now we can evaluate this log and what i'm going to do is i'm going to make that two bigger the 3 comes over and it's the power on 2 and the x squared minus 1 goes to the other side it's the same thing every time and from here we solve for x so 2 cubed is 8 now we get eight is equal to x squared minus one and from here there's a few ways that you could solve for x you see the x squared here and so i'm assuming you're probably tempted to move that eight over and get zero is equal to x squared minus nine and you can do that that's completely fine but another thing that you can do here if we don't have an x term in our quadratic what you can do is you can add one to both sides like get everything that's not an x squared to the other side and then you can just square root both sides and if you do that you get x is equal to a plus or minus 3. remember that when we square root both sides like that we have to put a plus or minus on one of the sides so we get that that looks like it's going to be our solution but we need to double check will first off positive three work in both of these arguments will it give us a positive number each time one x plus one if we plug in three we get three plus one that's four okay that's positive and in x minus 1 3 minus 1 that's 2. okay that's positive as well so 3 is definitely a solution that works here but remember our other solution is negative 3. so if we plug in negative three right away you can see we get negative three plus one inside the first log that's a negative number so no we can't have that and so the solution isn't plus or minus three x equals 3 is our only solution x equals negative 3 not so much so we can say x equals negative 3 is extraneous all right moving on to the last problem for this video here what kind of problem do we have well i see two of our terms are logs but this two is not and so again what is our goal our goal is to get one log on one side and everything else on the other side but what's the issue here we have a log on both sides so what do we do well to start that process of only having one log on one side we have to get all of our logs to the same side and so what we can do here is subtract the log of x minus 1 on both sides when we do that we get the log of x plus 10 minus the log of x minus 1 is equal to 2. and now these cancel off and great so now that we have all of our logs to the same side we can combine them into one log these logs are being subtracted so this is a little bit different than what we've been doing we're still going to combine it all into one log the only difference is since they're being subtracted we're going to divide the arguments this time and so we get x plus 10 divided by x minus 1 and that is equal to 2. and from here we can evaluate this log now there's one issue here this log doesn't have a base does it i mean there's nothing here and remember that when a log doesn't have a base the base is 10. that's the default base for logs and so now we have everything that we need to evaluate this the 10 gets bigger 2 goes over it's the power on the 10 and the x plus 10 over x minus 1 goes over to the other side now we know that 10 squared that's 100 so we get 100 is equal to x plus 10 over x minus 1. and now where the heck do we go from here because you know all our x's are on one side yeah but we have this fraction here so that's kind of stopping us from doing anything so when we have that fraction our first goal is to get rid of that fraction it doesn't matter if the x's wind up on both sides and everything we can always move those back later the first goal is to get rid of the fraction and how do we get rid of that fraction well we can multiply by that denominator on both sides and yes if we multiply by that x minus 1 on both sides it is going to now give us x's on both sides but that's okay there's going to be ways to move it back later so this is kind of messy here but we can see that we have 100 times x then we have 100 times negative 1 and on the other side we have x plus 10. and from here you can see it's pretty quick to solve we can get all our x's to one side by just doing addition or subtraction here here we're just going to subtract x on both sides and 100 minus x is 99x now we got all our x's to one side but the negative 100 is still over there we can't have that so to move it over to the other side we add 100 and that gives us a 110 on the other side so now we divide by 99 on both sides to fully solve for x i know the numbers look a little gross but x equals 110 over 99 can be simplified because 99 that's a multiple of 11. that's 11 times 9 and well 110 that's 11 times 10. so if we divide this by 11 on top and bottom we can get that x is equal to 10 over 9. so that's great we get our solution but we also need to check this and make sure it's not going to make any of our logs negative in the original problem now our solution is 10 over 9. so let's look here 10 over 9 that's a little more than one so we have something that's a little more than one plus 10 that's going to be a positive number and then we have something that's a little greater than one minus one that's still going to be positive so 10 over 9 is going to work here and so that is our solution to the last problem for this video and so that is log equations in a nutshell and if you're feeling pretty good with this then here's a problem for you to try and answer in the comments here we have log of x plus log of x plus 5 is equal to log of 6. so give that a shot let me know what your answer is in the comments and if you have any questions on anything we talked about in this video again let me know in the comments and i'll try to get back to you when i can now remember i do have that extra video where you and i will go through and solve 12 more of these log equations so if you're really looking for that extra practice like if you have a quiz or test coming up on this kind of stuff then i highly recommend you check out that video which i have linked in the description lastly make sure that you guys are subscribed to the youtube channel we're getting closer to 100k by the day and i can't thank you enough for you guys to support it's really it really helps get through all these videos and everything because you know at times it does it does take a lot of like mental effort to keep going with all this so anyways that's gonna do for this video and i'll see you guys soon