 So let's try to solve exponential equations. And in general, we can use the rules of logarithms to simplify equations involving variable powers. So note that this is not a case where we have a variable raised to a power x raised to the power 3, but rather a number is raised to a variable power 5 is raised to the power x. So what I want to do is I want to first transform my equation so that I have a product quotient power root or number equal to a product quotient power root or number. And the reason that this is important is that you have a way of simplifying the log of a product quotient power or root and the log of a number is just a number. And what this means is that if I have product quotient power root number equal to product quotient power root number, then I can hit both sides with the log. And for now, at the beginning, we'll leave the base unspecified. I'll use the rules of logarithms to simplify the expression that I have. And if it's possible to express all of the remaining arguments as integer powers of some base, we'll go ahead and use that base and then simplify to complete our solution. Otherwise, and this is what happens more generally, if you can't do that, we're going to use that special base, E, and be working with natural logs. So for example, let's take the equation 3 times 16 raised to the power 3x equals to 24. So a little analysis goes a long way. On the left-hand side, I have 3 times 16 raised to the power 3x. This thing over on the left is a product equal to number. And because it's product equal to number, I can hit both sides with the log and use the rules of logs to simplify. So I'll take the log of both sides base unspecified. I have the log of a product. So on the left-hand side, the log of a product is the product of the logs. And the second term. So if I look at this, I have a sum. The first term is a log of a number. And so that's just a number. The second term is the log of a power. And I can apply the power rule. At the exponent, the power becomes a multiplier. And so now I have this expression, simplifying log 3 plus 3x log 16. Now I have an equation, and I want to solve this equation for x. So what do we have? On the left, we have a sum. So the first thing I'll do is I'll subtract. And on the right-hand side, I have log minus log. This is the log of a quotient. And so this will be the log of 24 divided by 3. So I can simplify that. And that's a nice, simple arithmetic value that I can compute. That's going to be the log of 8. So solving for x, I get x is log 8 over 3 times log 16. And there is my initial solution for x. Again, I haven't specified what the base is. So I can leave it in this form if there's no obvious base. Actually, if I want to change the form later on, I'll use the ln version of log. But I do want to see if I can express the remaining arguments, the things I'm taking the log of 8 and 16. I want to see if I can express those as powers of some base. And so the thing that I might notice is that 8 is 2 to the third power and 16 is 2 to the fourth power. And so both 8 and 16 are powers of 2. And because of that, I know that a good choice of base is going to be square root of e raised to pi over 4. No, wait. They're both powers of 2, so I should use 2 as the base of my logs. So here, I'll go ahead and fill that in. x is log to base 2 of 8. And likewise. And this allows us to simplify the expression because I've already made this observation here. I know that the log to base 2 of 8 is the exponent 3. The log to base 2 of 16 is the exponent 4. And by substituting those in, I can simplify my expression. And I can do a little bit of basic arithmetic to simplify this expression to my final answer, x equal to 1 quarter. As always, we should check our solution just to make sure that everything was done correctly and also that we haven't introduced an extraneous solution. So the question is, if I let x be 1 quarter, do I actually get 24? So I'll substitute that in. I'll take care of the right-hand side. There it is. Over on the left-hand side, I have this is a product and I can treat this as a power raised to a power. 16 to power 1 quarter is 2. And so this is going to be 2 to the third times 3. 3 times 2 to the third. It's a true statement. So our solution is going to be x equal to 1 quarter. So here's another exponential equation. And a little analysis goes a long way. And in this particular case, it keeps us from making the wrong move. Because we have a sum over on the left-hand side, we have a sum we cannot simplify this using logarithms. There is no rule that simplifies the log of a sum. Fortunately, we do know algebra so we can transform the sum into a different thing. And rather than having a sum, I can subtract 8 from both sides. I can eliminate an add-end. And well, I do have to calculate 53 minus 8. So that's going to be 45. And now I have, over on the left-hand side, 15 times e to the power negative 0.01x. This is now a product and that means that I can use the rules of logs to simplify. I'll hit both sides with the log and I can use the rules of logs to simplify the product. Log of a product is the sum of the two logs. I can use the rules of exponents to not do anything with that because I don't know what the value is. But this is e raised to sum power and so I have a rule for a log that helps me simplify powers. This exponent comes out front and I have my logarithmic equation. Now I can solve this equation for x. So let's see. This is a difference. So I'll get rid of the other term. That's a difference of logs so that simplifies as a quotient and 45 over 15. I can simplify that as well. And let's see. I want to disentangle all the other stuff from the x. This is negative 0.01 times x times log e and because it's a product I can get rid of the other factors by division and again x equals log 3 over all the rest of them. And so here's my initial solution for x and I just have to pick a base. So the remaining arguments are going to be 3 and e and neither of these is obviously the power of sum base. So we'll use base e and simplify where we can. So first of all, whenever we're using base e we do want to use the function ln, not log so log to base e of 3 we can write as log 3 by the way, ln is correctly pronounced log. There is only one real log. And let's see, what else do I have? Well I know e to power 1 is equal to e. So that says that log base e of e itself is the exponent is equal to 1. So this part here is 1. So this is ln 3 over negative 0.01 which I'll simplify a little bit further and there's my value of x.