 So let's see what we can do with evaluating logarithmic expressions. And so all of these go back to the same basic definition of logs and exponents, which is, again, if I ever have a to the power n equals b, I can immediately write log to base a of b is equal to n. And so these two things mean exactly the same thing. If I ever see one, I can immediately write down the other. So this also means that the exponent is the log. And so many of the rules of exponents can apply also to the rules of logarithms. So let's take a look at this. This is log of, say, we want to find log to base 5 of 25 times 125. And so I'll give a name to this. I'll let x be this thing that I want to find. And importantly, I want to use the definition of logarithms to rewrite this equation. So let x be log to base 5 of 25 by 125. And my definition, a to the power n equals b if and only if log to base a of b is equal to n. So I'll compare my logarithmic expression with what I have. And that tells me my base is 5. Log to base a, log to base 5, a is 5. b, the thing I'm taking is the log of the argument 25 times 125. And n, the value of the log, is going to be my unknown x. So I can rewrite my expression 5, that's my base, to power x. The exponent equals 25 times 125. So again, a is 5, that's my base. My exponent is n, my exponent is x, and my power, the thing that is equal to the exponential expression, is 25 times 125 is equal to this. So now what do I know? Well, I can go over the powers of 5 to make the comparison easier. I want to rewrite everything so that I have an exponential expression with the same base. So 5 squared is 25, and 5 cubed is 125, so I'll use that. And I do have the rules of exponents. I'm multiplying two things with the same base, so I'll add the exponents. 5 to power 2 plus 3, I'll do the hard work of the arithmetic, that's 5 to the fifth power. And so that tells me, comparing my two expressions, 5 to power x is the same as 5 to power 5. And that tells me that x and 5 must be the same thing. And so that x is 5, and again, since x is log to base 5 to 25 times 125, then I know this is going to be equal to 5. How about a different one, log to base 4 of 4 over 32? So again, I'll let x be log to base 2 of 4 over 32, and I'll use my definition of logs to rewrite the equation. So again, my definition of logs, a to power n equals b, gives me log to base a of b equals n and vice versa. So a is my base is 2, b, my argument, 4 over 32, and the thing the log is equal to is x. And so now I can write this, a to power n equals b. That's a to the power n equals b, 2 to the power x equals 4 over 32. From this, I have this. And I can now write everything as powers of 2. So 4 is 2 to the second 32. I can figure that out. It's going to be 2 to the power 5. And so I have 2 to the x equals 2 to the 2 over 2 to the 5. I can apply the rules of exponents, 2 to the power 2 minus 5. And that gives me 2 to the power negative 3. And again, a comparison of the two expressions, 2 to the power x, 2 to the power negative 3. They are the same thing. They are equal. So that tells me x and negative 3 have to be the same. And again, x is log to base 2 of 4 over 32. So that tells me what log to base 2 of 4 over 32 is. Well, since this is base 10, we have to use differential calculus to expand the right-hand side using a power. Well, actually, we can do this exactly the same way we've done all the other logarithmic problems. This is base 10. It doesn't matter what the base is. I can let x be this expression. And I want to use my definition of logarithms to rewrite the equation. So base is 10. Argument 10 to the power 500. x is the log, is the exponent. So that gives me 10 to the power x equals 10 to the power 100 to the power 50. That's 10 to the second to the power 50. And I'll use my rules of exponents again. That's 2 times 50. That's going to be 100. And so that tells me that x is 100. x is the log. And so I have this value.