 This video is going to show how to use logarithms to solve equations, and they're coming very handy when we're trying to solve exponential equations, especially if they're more involved in problems that we've seen previously. I mean, we can always use our calculator to intersect, try to or use our table and try to find an approximate solution, but if we want an exact answer, then we're going to need a logarithm. So there's this property that we have in logarithms that helps us be able to solve these equations. So let's take a look and see if we can figure out what this property really tells us. We want to put these into our calculator just like we see them. So log of seven, care at three, and then close the parentheses, and we find out that that is 2.54, if we approximate it, so approximately 2.54. And if we put in three times log of just seven, we get that same thing, 2.54, approximately. Now let's look at this one. Natural log of six, care at negative two, close the parentheses, and we find out that that is approximately negative 3.58. And if we do negative two and then ln six, we get that same thing, negative 3.58. So what do we notice? We notice that in this first two, they both had logs, they both had sevens, and they both had threes. It's just the three that changed. The three was in the exponent over here, and it was a coefficient in the second choice. Same thing happened over here with this natural log. We had negative two as an exponent in the first one, and then that negative two is our coefficient, and now our six in our, inside our parentheses has no exponent on it. And it looks like the exponent went in front of the log, which is exactly what the power property is. The power property says log base b of x to the n. There's an exponent inside the parentheses. And that's this equivalent to n, that exponent, times the log of base b of just the x. So the exponent comes out front. So how do we use that? Well, we want to isolate the exponential, so we want to get that base all by itself, and then move any extra pieces around so that we can get b to the x by itself. And then we want to apply one of our logarithms on our calculator, either the log or the natural log, to both sides. Then we can use that power property to solve for x. And then we can always check our answer by plugging it back in or using the calculator graph or table. So let's look at this first one. Four to the x. We already have it isolated. The base is, with its exponent, it's already isolated by itself, so I'm ready to take the log or the natural log. This time I'll take a log, but it really doesn't matter if we take the log or the natural log, but give us the same answer. And then I took the log of four to the x, so I have to take the log of six. That's what it means by taking the log of both sides. And now over here on this side, my exponent is in part of my argument, the log that I'm trying to take. So I can bring it out front, and it becomes x times the log of four. I've moved that x, so I don't have to write it twice. I'm just writing it in a different place. Equal to the log of six. And if I want to solve for x, this is really that log times that x, so I would divide off the log of four, and divide off the log of four on this side, or divide by the log of four. So x is going to be equal to the log of six divided by the log of four. Now that is an exact answer. You can leave it at that. This is a perfectly fine answer. This little statement up here says, remember that the log is just a number, so if you don't want to see that log, if that just bothers you and you don't read, it's harder for you to recognize that as a number, then we can just, we can find the decimal. So we take log of six, exactly like we see it. So I have to have the log of six divided by, and then I have to write log again, of four. And when we do that, we get one point two nine, approximately. So either one, either one works. And if we just want to double check for ourselves, we can come in here and say four to the, if you want to be exact, we would say four to the log six divided by log four, four carat, and then parentheses, because we've got a division, and literally what I see, log of six, and then I close that argument, so then divided by the log of four, and close that argument, and then close the exponent, and I see that that's equal to six. Now if I had like the decimal, when I went four to the one point two nine, it's going to be approximately six. Let's watch. Four carat one point two nine, and it's five point nine seven nine, nine seven nine, which is approximately six. Close enough. So one point two nine works, but if you want the exact answer to come up, then you need to keep it with the log. So here's where I'm headed. I want to get this by itself. That means I need to take the 21 to the other side. So six to the x is going to be on this side, and then 21 added to three will be 24. Now don't mistakenly think that six times four is 24, that x is four. Remember this is an exponent. Six to the something, okay? We know that six to the first is six, and we know that six squared is 36. So our answer is actually going to be one point something, because it's in between. So let's try, and see if we really get that kind of an answer. That's how you can intuitively think about it. So we take the log of both sides, and this side just to be different, I'm going to take the natural log. So the natural log of six to the x is equal to the natural log of 24. And I should put these in parentheses. They don't have to be in parentheses, but when you go to put them in your calculator, you won't forget if you keep carrying the parentheses. All right, the x has to go out in front, because the property says that when I haven't explained it in my argument, I can multiply it times the log and take it out of the argument, because it's an exponent times an exponent. That's what we do when we raise a power to a power. So x times ln of six is equal to ln of 24. And I divide off the ln six, so I can get x by itself. So I divide both sides by ln six. And x is going to be equal to ln 24 divided by ln six. Okay, what do we do here? Well, we got it again. We're headed here. So this is the farthest thing away, because it's plus 10. This is right next to it, because it's being multiplied. So we're going to take this first. So we still have negative seven times 0.8 to the x is equal to subtracting 10 will be minus eight. Then we're going to divide both sides by that negative seven. We're still trying to just get to the base by itself. That's all we're trying to do at this point. So 0.8 or just 0.8 to the x is equal to a negative divided by a negative is a positive eight. Divided by seven. And I would leave it that way, because we can use parentheses when we take the log of both sides. So don't simplify here. Just leave it just like that. It's so much easier. And we want an exact answer. So we don't want to round anywhere that we don't have to. So take the x out front. x times the log of 0.8 over the x one out front. We don't have to write it again. Equal to the log of eight over seven. So we don't have the exact answer here. Divide both sides by the log of 0.8. So x, I forgot my parentheses, is equal to log of eight over seven divided by the log of 0.8. Which is approximately, but you don't have to say it. We're good here. But if, again, the log doesn't mean number to you, then you can come into your calculator and say the log of, and it opens the parentheses for me, so that's why the fraction isn't hard to use at all. Eight divided by seven close my parentheses, then divided by log, just like I see it, of 0.8 close the parentheses. And it's approximately equal to negative 0.598. And I think I'll go out that many decimal places. You could say approximately negative 0.6 if you round me. And then our last problem here. Now it happens. I'm trying to get to the base all by itself while it is. But look at that exponent. Kind of an interesting exponent. It's not just x or it's not just, you know, it's got terms in my exponent. Okay, property still works. You just have to remember that this x minus seven has to work all together. So rewriting the problem, t minus seven is my exponent, so I'm going to write t minus seven. And I'm actually going to put it in parentheses because it's one exponent. It's one piece, so it's in parentheses, groups it together. And then times that log of three equal to that log of 65. There's nothing to do over here. It's just a log of a number. Divide the log of three so we can get the exponent by itself. So t minus seven. And I really can drop the parentheses now because it's the only thing left on that side. Is equal to the log of 65 divided by the log of three. And the only thing I have to do is get this seven to the other side. T is equal to the log of 65 divided by the log of three. And remember, that is a number. So I have this number that I want to add seven to. I have to add seven to both sides. And this would be my final answer.