 In this video, we're going to talk about solving exponential equations. This is something we've approached before in this lecture series, for which we use the fact that if you have, like, a to the x is equal to a to the y, that is, if you have these two exponential expressions that are equal to each other, because exponentials are one-to-one, we get that the exponent's x would have to equal y. That's a very common tool you can use to solve exponential equations. But what happens when the right-hand side becomes crazy almost seems like impossible? It's not impossible. But if it becomes crazy hard to write one of the sides of the equation as an exponential, like, I mean, how do you write 5 as a power of 2? That's a little bit more challenging. So instead of using this one-to-one approach, the idea here is, well, you could switch it from exponential form to logarithmic form, right? So you have this equation 2x equals 5, you have this base 2, because you want to get x all by itself on the left-hand side. You want to move this exponential base 2 to the other side, so you're going to switch it to its inverse function. So exponential base 2 turns into the logarithm base 2 of 5. And so that would be the answer to this question here, log base 2 of 5, because the log base 2 of 5, what that is computing here, this number is the power of 2 that gives you 5, which is exactly what we need to figure out right here. We need the power of 2, which gives you 5. Now, you probably want to estimate that. You might need to, let's say you need to round it to 3 decimal places or something. It's very likely when you consult your calculator, it doesn't have a log base 2 button or a general log button. So by the change of base form, you can write this as the natural log of 5 divided by the natural log of 2. You can use the common log also if you prefer, but in calculus, you see that the natural log really is the superior log. I mean, there's a reason why we call it natural. The natural log is the naturally occurring logarithm, believe it or not. And as such, I'd say for college hours, we get in the habit of using the natural log right now. You can approximate the natural log of 5 on your calculator, the natural log of 2. While those are both in your memory bank of the calculator, take their quotient. I noticed earlier, I wrote that it was approximate. These things are equal so far. The natural log of 5 divided by the natural log of 2 is equal to log base 2 of 5. But this, when you consult your calculator, you get about 2.322. So those would give you, that is a solution to this one. And that's an appropriate approach to take to this one. I do want to present an alternative approach, right? When you take something like 2x to the 5 right here, another approach that one could take is that, you know, since you had to use the natural log to approximate your calculator, one sort of asks the natural, we have the following thought. What if we just start with taking the natural log of both sides, right? If you take the natural log, you know, as long as you do the same thing to both sides of the equation, what's good for the goose is good for the gander, you're going to get the equation solved, right? Equality will be preserved as long as you go along. And so if we take the natural log of both sides, the nice thing is when you take the natural log of the left-hand side, the third law of logarithms comes into play, right? Exponents inside the logarithm come out as coefficients. And so this becomes x times the natural log of 2 is equal to the natural log of 5. And when you're working with logarithms, an important thing to remember here is that logarithms are just numbers, right? When you look at something like the natural log of 2, that's just a number. A rational number, mind you, but it's just a number. And so if you had something like 2x equals 5, you know exactly what to do. You divide both sides by 2. Well, in this situation, our numbers just have these, you know, there are rational numbers which we're using logarithms to describe them. This is just the natural log of 2 times x is equal to the natural log of 5. It's just a number still, the natural log of 2. So divide both sides by the natural log of 2 so that they cancel on the left-hand side and you get that x equals the natural log of 5 over the natural log of 2, which we saw earlier, kachink, that's the same answer, right? This is just log base 2 of 5. So we got the same answer in a slightly different way, right? And this one seemed like it was a little bit longer process. It's like, well, if you switch to the exponential form you got there immediately, why did you take the natural log of both sides? Well, we're going to see that for more challenging equations, starting off by taking the natural log of both sides is actually a very fruitful strategy. Let's look at a slightly more complicated example. Let's solve the equation 8 equals 3 to the x equals 5. Well, one approach would simply just be to start ripping off everything attached to the x here. So we're going to start off by dividing both sides by 8. So we get 3 to the x equals 5 eighths, like so. And then like we said last time, so what we were trying to do, why do we divide by 8? We're moving the 8 to the other side. We were multiplying by 8. So to move to the other side, you get its inverse operation division by 8. It's the same thing here. We want to move the base 3 to the other side and that's going to give us x equals the log base 3 of 5 over 8, like so. For which, again, if you want to approximate this thing, you can soldier calculator. At the very least, you're going to have to write this probably as the natural log of 5 eighths over the natural log of 3. I should also mention that by the change, by the laws of logarithms number 2 in particular, if you have a fraction inside of a logarithm, that's the same thing as taking a difference of logs. This is the log base 5, excuse me, the natural log of 5 minus the natural log of 8 over the natural log of 3. And so either one of these ones, all three of these things are all the same thing. It doesn't matter which one you use necessarily, but you can soldier calculator. You're going to get negative 0.428. So that's our solution right there. If you want to solve it using this exponential form. But like we saw earlier in the previous example, I claim we can actually solve this very effectively just by taking the natural log of both sides, right? Take the natural log on the left. We take the natural log on the right. And what do we get here? Well, on the left hand side, we have a product of two, two terms 8 times 3 to the x. When you have a natural log of a product, this becomes a sum of natural logs. You get natural log of 8 plus the natural log of 3 to the x equals the natural log of 5. Okay. So then you want to move the natural log of 8 to the other side. It's just a number. So if you're adding the natural log of 8, we'll have to subtract it from both sides. And so we end up with the natural log of 3 to the x is equal to the natural log of 5 minus the natural log of 8. That seems like I've seen that somewhere before, right? Then, like we saw in the previous example, you can use the third law of logarithms to pull out the x. And so you get x times the natural log of 3 is equal to the natural log of 5 minus the natural log of 8. And then you divide both sides by the natural log of 3. And looking there, you get the exact same number. X is equal to the natural log of 5 minus the natural log of 8 divided by the natural log of 3. It's the exact same number. The process is a little bit different. It is. But because of the laws of logarithms or the laws of exponents, whichever you need to use at the time, they're going to be equivalent to each other because if there's two correct ways to solve an equation, guess what? It doesn't matter, right? If they're both right ways, then they'll give you correct answers when you're done. And so that's so you have a little bit of variety on how to solve these things. And like I said, the more complicated the exponential gets, the more this yellow method turns out to be very fruitful. So for example, look at this one right here. Let's solve the equation 5 to the x minus 2 is equal to 3 to the 3x plus 2. In this situation, you'll notice that you have an exponential base 5 on the left-hand side and you have an exponential base 3 on the right-hand side. Why in the world would you ever consider such a thing? Well, first of all, notice if you had exponential equations f of x equals 5x minus 2. So this is an exponential base 5, which has been shifted to the right by two units. If you had g of x, then s3 to the 3x plus 2, right? So this one's an exponential function base 3 where there's been some horizontal compression and shifting and such. And if you want to find the intersection, right, where are these things intersect each other, right? So their graphs, you might see something like the following, like one could look like this, and then the other could look something like this, where there's a point of intersection. Where is that intersection? You'd have to solve the equation f of x equals g of x. So that's the exact situation we'd be looking at something like this. So this is not such an unnatural equation. This is a very natural thing to consider when you are looking at exponential growth and things like that. So how'd you solve this one? Well, my recommendation is that as opposed to the previous case, where you have this mismatch of bases, like the base 5 and base 3, so we should just be like, eh, I don't care. I'm just going to take the natural log of both sides, right? So let's stop trying to compromise on what the base should be. Let's just take it as it is and just take the natural log of both sides. I'm going to take the natural log on the left. I'm going to take the natural log on the right. Because in this situation, it doesn't really matter what the base is, because the natural logs, you can still use logarithms to pull out these exponents. So you're going to get the natural log of 5 to the x minus 2 on the left-hand side by the third law of logarithms. We can pull that exponent out, and we end up with x minus 2 times the natural log of 5, which remember, the natural log of 5 is just a number, right? It's really no different than a 3, right? The next thing, you're going to have the natural log of 3 on the right-hand side raised to the 3x plus 2 power. You can pull that exponent out by the third law, and so you get 3x plus 2 times the natural log of 3. So again, these are just numbers, right? It'd be like if we have a 7 and an 8 or whatever, you can distribute these numbers, okay? For which on the left-hand side, you'll get x times the natural log of 5 minus 2 times the natural log of 5. On the right-hand side, we're going to get 3 times the natural log of 3 times x. That's just the coefficient of x, and then you're going to get a 2 times the natural log of 3. These are just our coefficients right here. We want to combine like terms, so we're going to move this one to the other side, and we're going to move this one to the right-hand side. So we end up with the natural log of 5x minus 3 times the natural log of 3x, okay? On the right-hand side, we get 2 times the natural log of 3 plus 2 times the natural log of 5, all right? And so then on the left-hand side, you'll notice that we have like terms we can add together, or in this case, subtract their coefficients, or another way of thinking about it, you could factor out the common divisor of x right there. So the left-hand side becomes the natural log of 5 minus 3 times the natural log of 3 times x. Let's see, on the right-hand side, I mean, you could factor out the 2 if you want. I mean, it's not super helpful, but you know, when you see a common divisor, I just can't help myself. I have to factor it. And then to get x all by itself, we're going to divide by its coefficient, which is the natural log of 5 minus 3 times the natural log of 3. We do that on the right-hand side as well. In which case, then, we have our solution. x equals 2 times the natural log of 3 plus the natural log of 5 over the natural log of 5 minus 3 times the natural log of 3. There you go, for which you could throw that into a calculator and you would get approximately negative 3.212, which is honestly a perfectly good-looking solution right there. Now, if you had tried to solve this using some different approaches, you might get different things, but I want you to be aware that if you wanted to, you could condense this logarithm, this logarithm expression all into one single logarithm, right? How could you do that? Well, in the top, right, you have a sum of two logarithms, so that would become 2 times the natural log of 3 times 5, which would be 15. In the denominator, right, you have this negative 3 right here. You could bring it inside. So you get the natural log of 5 minus the natural log of 3 cubed, which is 27, okay? Keep on going. You could bring the 2 inside on the top, and so you end up with the natural log of 15 squared, excuse me. And then in the denominator, because you have a difference of logarithms, you could then write that as the natural log of 5 over 27. So let's note here, of course, that 15 squared is 225. So you get the natural log of 225 divided by the natural log of 5 over 27, for which, you know, if you wrote that as a single log, because this is a change of base type formula, your answer turns out to be log base 5 over 27 of 225, right? That's the same number, right? Again, this looks like a little bit confusing here, but when you look at the original equations, like, isn't it obvious? Oh, yeah, you should have been working base, you should have been working log base 527s, right? I mean, you can see kind of a retrospect. It's like, okay, I can see where the 5 came from. I see where the 5 came from, and the 27 was 3 cubed right there. You can kind of see in retrospect, but what I'm trying to say is when you're working with log or with exponential equations, you can kind of just forego any consideration of the bases and just switch over to natural log, because the natural log has all the capacities you need to solve the exponential equations. So don't necessarily worry about the base, just use the natural log. The natural log is your friend, I can promise you that.