LogarithmLogarithmicLogLog equationEquation

Hi guys! I'm Nancy and I'm going to show you how to solve logarithmic equations. In my other logarithms video I warned you that solving log equations could be a whole other video, so here we are. Let me show you how to solve them.

OK, so a quick recap, in my other logarithms video I showed you some simple log equations that just had one log in them and equaled a number. OK, so every log is connected to an exponential form. Logs and exponentials are opposite operations of each other, and the best way to figure out a log is to rewrite it into exponential form, and solve for X.

And in my other video, "Logarithms... How?", I showed you how to rearrange the log into exponential form, and solve for X. Basically, it's just the little base number, raised to the power of the number on the right side, equals the middle number, the argument or the number that's left, so... base to the right side equals the middle, base to the right side equals middle. If you don't remember how to do this you can jump to my other video, "Logarithms... How?" There was talk of memory tricks, like little to the right equals middle. There was talk of snails with counter clockwise spiraling shells, it was real party.

But basically this is the simple kind of log equations. What if it's not so simple? What if it's more complicated? You say, like more that one log, a log equals another log. Several logs equal a number. Many logs, some of them with numbers multiplied out front.. What do you do? And what do you do if you see something like "check your solutions?" Or if your teacher says, "Check for extraneous solutions" or "check your answers." How do you do that? I'll show you. And I'll also show you the log properties, which are what you need to simplify and condense, and solve more complicated equations like that. So that would be like the power property, quotient property, product property, equality property, all of those. So my other video gave you a taste of solving log equations at the end, and now I'm going to give you a whole meal.

But if you see the kind that you're working on you can skip ahead. So just check the description for the time to jump to. Let me show you how to solve. Alright, say you have a log on each side of the equation. It's getting more complicated, but don't worry, don't panic, because actually really simple, if the logs have the same base you can just set the insides of the logs equal to each and solve. So this is the equality property and it's just like I was saying, if the little bases are the same you can take the insides of the log and set them equal to each other. By the way this is for little bases that are positive and not equal to one.

But anyway, if the bases are the same, you can set the insides equal, and I know what you're thinking, "What bases?" "I don't see any bases." Yeah, they're not explicitly written here. If you don't actually see them, it's implied to be 10, the common log. So that's like the default, and we can just go ahead and write that. And now that we can see that those bases are actually the same. You really do just take the inside X expressions set them equal to each other, and a new equation, and leave the logs behind. So then if you do some algebra and solve, you get x = 2.

And I know you might think that you're done, but with these log equations, you actually have to check the numbers you get. So take the number and plug it back into the original log equation and make sure everything checks out. The reason is that sometimes the algebra you do after leaving the logs behind can magically turn up extra numbers that don't actually work for the log.. for the domain of the log. Let's try it. So you take that number for X, and plug it in for X in the original log, and make sure everything checks out. All right when you plug in 2 for X, you end up getting log of 17 = log of 17, which is true.

But also, log of 17 is perfectly fine, because that's in the domain of the log. By the way if you ever got log of a negative number, you would have to throw out that answer, because log of a negative number is undefined in the real number system for the domain of the log. But our's is fine, our 2 is fine. It checks out so we can box that. And that's our answer. So that's the equality property, if you have one log equals another. Now let me show you the product property.

Alright, what if you have two logs on one side of equation, and then just a number on the other? Don't worry, you're going to use a property. And a lot of the work of these equations is taking the multiple logs and combining them into one. And with something like this, two logs added together is going to end up being the same as one log of a product. Let me show you what I mean. OK, so this is the product property, and all it says is that the sum of two logs is the log of a product. In other words, addition outside two logs is the same thing as multiplication inside a log. I'll show you what I mean.

And by the way this is for variables that are positive, and when the base is not equal to one, but anyway what does that mean for us? For our example? It means that because these bases are the same, both have base 2. Instead of having two logs added together on the left, we can write it as one log of these two multiplied. One log of X times (X - 2), the log of their insides, their arguments multiplied. Alright so we've rewritten it as one log. So we have just log base 2 of X(X - 2), and it still equals 3 on the right. This is great, because we just have one log in the equation now.

And we can use what you know from before to solve by rewriting it into exponential form. So that would look like base 2 raised to the power of 3 equals this part X(X - 2), that's how you rearrange it into exponential form. Alright, so here's all the algebra we do to solve after you write it in exponential form. You want to get everything on one side, and just 0 on the other side so that you can solve. This quadratic you can factor into X - 4, X + 2. And once you have it in factored form equals 0, the way you solve is setting each factor equal to 0.

And we get X = 4, and X = -2. It's tempting to think that you're done, I know. But again we have to take those numbers, and check them in the original equation.

And I know, I know I didn't write check your solutions this time. I left it out on purpose because your teachers will do that sometimes.. to really test you. They're trying to trick you, it's shady, I know. But we have to check those numbers, so we'll take 4 and -2 and plug them each in for X in the original equation. Alright, so we checked both answers.

And we check 4 for X you end up getting this, log base 2 of 4, is just 2 because 2 to the 2 power is 4. That's just equal to two. This is just equal to 1, because 2 to the 1 power is 2. Anyway on the left side it's really 2 + 1, which is equal to 3.

So that checks out, and it means that 4 is an actual answer. Now -2, when we check that one we end up getting logs of negative numbers. And anytime you get a log of a negative number it's undefined in the real number system. The negative number is not in the domain of the log. So whatever caused that to happen has to be thrown out as an answer, so it's not really an answer. And the only actual answer is X = 4. So that's how you use the product property if you have two separate logs added and they have the same base, you can put them together as one log of the two insides multiplied.

Now let me show you an even more complicated equation where you need the quotient property and the power property. Oh, but it gets better guys! Now we have four logs. And one of them has a number multiplied in the front, what a mess. Yeah, you can't un-see that. But it's actually not that bad, because the first thing you want to do is anytime you do see one of those numbers multiplied in front of a log, you use this new property, the power property, and you bring it up as a power inside the log. OK, so this is the power property, and all it says that if you have a number multiplied in front of a log, you can bring it up as a power inside the log. And this is true for real powers P and when M and B are positive, and B not equal to 1. That's my disclaimer.

But anyway, for our problem, because we have this number multiplied in front, the 2, as a coefficient, we can use the power property and bring that 2 up as a power here for the 3, inside the log. So this term will become log base 4 of 3 to the 2 power, 3 squared. And of course 3 squared is just 9, so we can replace that with 9. OK, now it's starting to look more like something we've seen before, kind of. Try to use what you already know, and if you see two logs added together and they have the same base, think of that property from before.

the product property where you can combine them into one log that is of the product of these two. So log base 4 of 9 times 2 combined into one log. So we combine those two logs together, became log base 4 of 9 times 2, which is a log base 4 of 18, and this is our current version of a log equation, it's getting more and more simple, more and more condensed, as they say, which is what you want. Really we want just one log on each side, max. So now looking at this, there's not much more you can do on the right. But on the left, you may already have guessed this but when you have two logs subtracted, one log minus another and they have the same base, there's a property for that. And it's not the product property but the quotient property.

So this is the quotient property, and all it says is that if you have one log subtracted from another, and they have the same base, you can rewrite it as a single log of one inside over the other inside. What that means for our problem is that on the left side we can write this as just one log. log base 4 of 36 over 2X. Alright, and now.. I wrote this in a way that hopefully it'll make it pop out at you, but.. at this point it should look somewhat familiar.

If you have just one log equals another log, and they have the same base, you can just take the insides and set them equal to each other and solve. Alright, so when you do the algebra and you bring the 2X over to the right and you solve you get x = 1. And I'm sure by now I don't have to remind you that you do need to check that number. Plug the 1 back in to the original equation, and make sure everything checks out.

OK, so this answer does check out. If you plug in 1, and you simplify, you can use all the same properties when you're simplifying. But this does check out, so x = 1 is the actual answer. And this was a good example, we used all four properties. Power, quotient, product, and equality. So these are all the properties you can use to solve log equations. And I should tell you that also, I know this video is on log equations, but if you just have expressions, log expressions that you're supposed to simplify, you can use those same properties and lot of the same ideas if you have to do that.

So I hope that helped you understand how to solve log equations. I know logarithms are exactly what you wanted to be doing right now. That's OK, you don't have to like math.

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