 This video we'll talk about solving logarithmic equations. So the most common, says on your paper, mistake is for people to do the inverse function too early. So we've looked at some problems where it looks something like this, 2x minus 5 is equal to 15. Well some people realize they're going to have to take a log here and they would want to take the log no matter what kind of log we want and do it now. That's too early. What do you have to do? You have to get this base by itself first. Your goal is to get to that by itself. We looked at logs before and we've had some log equations that look something like the log of x plus 1 minus 3 is equal to 10 and people want to say oh well I can just exponentiate that right away. Take the inverse function so exponentiate and they would write something like this and they would say well that's 10 to the 10 if you were really doing it, exponentiate incorrectly. But again this is too early. Why is it too early? Because we need to get to this log x plus 1 first. You've got to get to this 2 to the x first. Now we need to talk about the domain of logs because that's the other problem that people have. The domain of a log is that x, that argument, has to be greater than zero. In other words you can't take the log of a negative number. If we had log of negative 2 if we tried to put that in our calculator it would give us an error because you can't take the log of a negative number. So this would cause us, sometimes we get these extraneous roots. So there are solutions when we try to solve the problem that solve it according to our work but if I go back and plug it in it doesn't work. Let's go through this example. Log x plus log x plus 9 is equal to 1. We can apply a property here because we're adding until we can combine it into one log. So log of x times x is x squared and x plus 9 would be plus 9x and it's equal to 1. No properties on the other side. And this one I think I just converted instead of taking the base 10 of everything actually in here it should be 10 to the log x squared plus 9x is equal to 10 to the first. And then the log and the exponential can't switch other out so we're left with the x squared plus 9x is equal to 10 to the first. Bring 10 over x squared plus 9x minus 10 is a quadratic so we set it equal to 0 in factor. We get x plus 10 and x minus 1 and that gives us x equals negative 10 and 1. But when I go back and plug them in, log of negative 10 plus log of negative 10 plus 9 if I get to where I'm taking a log of a negative number anywhere then I can't have that solution and this is no good so we say 10 doesn't work and if we come back in here and say log of 1, well that's positive we can do that one. In fact we should know what that is and the log of 1 plus 9 well that's just going to be the log of 10 so we can say that this one works because everything was positive when we plugged it back in. So we can't forget to do that. So solve the equation, check your answer and these are going to be using our properties to get us started. So we need to use the property that says log of a plus log of b is equal to log of a times b. So we're going to have a single log and 5 times x minus 9 so it's going to be the log of 5x minus 45 and that's equal to 1, exponentiating we get 10 to the log 5x minus 45 is equal to 10 to the first or just 10 log and the exponential cancel each other out they undo each other probably the better way to say that so 5x minus 45 is equal to 10 it's linear so I want to add my 45 to both sides so x is equal to 55 and x is going to be equal to 11. So I put 11 in here where x is that's going to be 11 minus 9 that's where my only x is so 11 minus 9 is going to give me the log of 2 and we're good so we can say that x is equal to 11. Next example now we have a subtraction going on here so this is the one that says log base n minus log base n is equal to log of m divided by n so I take my single log and in my argument on top I write my 2x minus 5 that's my m and on the bottom I put my 78 and that's going to be equal to negative 1 I can have a log equal to a negative number I just can't take the log of a negative number so we should be okay here and we are going to exponentiate so that's 10 is equal to 10 to the negative 1 some of you are going to be able to skip this step and be able to go bypass this one and go straight to the argument 2x minus 5 over 78 is equal to 10 to the negative 1 or 1 over 10 so if I want to clear my fraction here to get to x I have to multiply both sides by 78 I'm going to go this way so 2x minus 5 is going to be equal to 78 divided by 10 when I multiply both sides by 78 and now I need to take and add my 5 and I'm going to add and 5 would happen to be 50 over 10 so we can simplify a little bit which is equal to 128 over 10 and then 2x must be equal to that 128 over 10 and if I divide by 2 I have 128 over 10 divided by 2 which is the same thing as multiplying by 1 half so it's 128 over 20 and at this point I would probably be just happy with that or give me the decimal the decimal happens to be 6.4 so add the equation and check your answer and we have now logs on both sides so what do we do about that? Well we like to keep the logs positive if at all possible so I'm going to take this log this way and the negative 1 over here I'm getting the logs on one side so I'm going to have log of x plus because I'm taking this negative log to the other side the log of x minus 9 is going to be equal to positive 1 when I take the one to the other side so the property I'm adding these logs okay this is log a plus log b is equal to log a times b so we have a single log and remember we're going to distribute since we're multiplying so of x squared minus 9x and that's equal to 1 and I'm going to try skipping the next step if I exponentiate 10 to this log I'm going to have x squared minus 9x and if I exponentiate over here I'm going to get 10 to the first or 10 I kind of did it cheating I didn't write it again I just wrote the x write those things as my exponents it's quadratic so you solve it by setting it equal to 0 so it'd be x minus 10 and x plus 1 to give us a negative 9 so this tells me that x is going to be equal to 10 and x is going to be equal to negative 1 if I check 10 I have log 10 that's positive so we're good plus log of 10 minus 9 which is also positive so that's okay and that's supposed to be equal to 1 and that all works and then we have negative 1 that we have to try so log of negative 1 and replacing my x and right now that doesn't fit our domain so we have to say that this one is extraneous so x equal 10 x equal negative 1 is extraneous so we have another property of logs that will kind of help us out solving these and we've really talked about this with exponentials so it shouldn't be real new it says if we have m and n and be not equal to 1 then m is equal to n if I have these logs and if we have m not equal to n so if they're not equal then the logs are not going to be equal to each other so equal basis imply this time equal arguments let's solve this equation now use our properties and we have log base 3 and then we have x plus 6 over x because we had a subtraction here is equal to log base 3 of 5 and what this is saying to us is if we have the same log and with the same base then we really just have x plus 6 over x that argument is equal to the 5 which is the argument on the other side and now we have an equation to solve x plus 6 will be equal to 5x when we clear the fraction and subtracting the x to the other side 6 will be equal to 4x which makes x equal to 6 4 or 3 halves and if we plug it back in log base 3 of 3 halves plus 6 that's going to be positive minus the log base 3 of 3 halves and that's positive and then we would have equal to 5 so let's look at our next problem solve using the appropriate methods now we got all kinds of things that we know how to do and answer in exact form and we can check for and identify any extraneous roots don't forget to check is what it's saying I have two logs that are being added so to make it a single log this time I do multiply 7 times x would be 7x and 7 times 4 would be negative 28 and that's equal to 2 so I can exponentiate 3 the log with this thing is equal to 3 squared and the log notice it's this time it was 3 not 10 because we had a base 3 down here but I exponentiated with the base of the same base of my log so those undo each other and we're left with 7x minus 28 on that side equal to 3 squared which is 9 solving the linear equation 9 plus 28 is going to give me 37 and dividing by 7 is going to be 37 over 7 and that would be an exact answer we haven't simplified we haven't rounded 37 over 7 so we check see if we have positive things going on like base 3 of 37 over 7 minus 4 and this is going to be bigger than 4 because 28 would be 4 so this is positive and then plus the log of base 3 7 which is already positive so we should be able to say that this one's going to really work so what do we have here well I have one log on one side equal to a constant so I can exponentiate so 10 to that is equal to 10 to the first the log in the exponential cancel each other out after I exponentiate so negative 6 minus x is equal to 10 don't forget to do the exponent over here if I add 60 both sides negative x will be equal to 16 and x will be equal to negative 16 does it really log of negative 6 minus my x which is negative 16 equal to 1 well if I add my 16 that's going to give me the log of 10 is equal to 1 and this is what I was saying before we really could check this first we have no more x's anymore and if I were to check this by converting I'd have 10 to the first is equal to 10 is the exponent on 10 1 that will get me 10 sure is and I had a positive number so I can say I have a solution of negative 16 so notice solutions can be negative they don't think just because x negative that is not going to work they can be negative but we can't take the log of a negative number so you have to check all right I've got a subtraction going here so I can write a single log and my single log is gonna have a fraction in it since I'm subtracting I want to exponentiate so that I can have the same base and that my exponents are the same and while I'm exponentiating my log and my base cancel each other out so clear the fraction then it equal to 0 and we can factor that and it looks like we would have x plus 7 x minus 2 would give us negative 14 and add to 5 now we have to go and check so I get negative 7 for x and then I get log of 7 is it log of negative 7 then worse I'm really kind of done with that one so let's try the next one so here we plug in 2 for x the log of 16 minus the log of 2 is equal to the log of 8 and let's just check this one too make it a single log over here so I have the log of 16 divided by 2 equal to the log of 8 and what is 16 divided by 2 it's 8 so we really did find the right answer and we would say the x equal negative 7 is an extraneous and this is my solution