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## Algebra 2

### Unit 8: Lesson 4

The change of base formula for logarithms

# Evaluating logarithms: change of base rule

Sal approximates log₅(100) by rewriting it as log(100)/log(5) using the change of base rule, then evaluates with a calculator. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base. And that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense, or how we can derive it. So the change of base formula just tells us that log-- let me do some colors here-- log base a of b is the exact same thing as log base x, where x is an arbitrary base of b, over log base, that same base, base x over a. And the reason why this is useful is that we can change our base. Here are our bases, a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying-- this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x, and e is obviously the number 2.71, keeps going on and on and on forever. Now let's apply it to this problem. We need to figure out the logarithm-- and I'll use colors-- base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing as log-- I'll make x 10-- log base 10 of 100 divided by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100-- what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2. So it simplifies to 2 over log base 10 of 5. And we can now use our calculator, because the log function on a calculator is log base 10. So let's get our calculator out. We want to clear this. 2 divided by-- When someone just writes log, they mean base 10. If they press LN, that means base e. So log without any other information is log base 10. So this is log base 10 of 5 is equal to 2 point-- and they want it to the nearest thousandth-- so 2.861. So this is approximately equal to 2.861. And we can verify it because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power. And this number is closer to 3 than it is to 2. Well, let's verify it. So if I take 5 to that power, and then let me type in-- let me just type in what we did to the nearest thousandth-- 5 to the 2.861. So I'm not putting in all of the digits. What do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good. That this is the power that I have to raise 5 to to get to 100. Now with that out of the way, let's actually think about why this property, why this thing right over here makes sense. So if I write log base a-- I'll try to be fair to the colors-- log base a of b. Let's say I set that to be equal to some number. Let's call that equal to c, or I could call it e for-- Well, I'll say that's equal to c. So that means that a to the c-th power is equal to b. This is an exponential way of writing this truth. This is a logarithmic way of writing this truth. This is equal to b. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this? 10 to the same power will be equal to this, because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base. And I'll actually do log base x to prove the general case, here. So I'm going to take log of base x of both sides of this. So this is log base x of a to the c power-- I try to be faithful to the colors-- is equal to log base x of b. And let me close it off with orange, as well. And we know from our logarithm properties, log of a to the c is the same thing as c times the logarithm of whatever base we are of a. And of course, this is going to be equal to log base x of b. Let me put-- I can just write a b, right over there. And if we wanted to solve for c, you just divide both sides by log base x of a. So you would get c is equal to-- and I'll stick to the color-- so it's log base x of b, which is this, over log base x of a. And this was what c was. c was log base a of b. It's equal to log base a of b. Let me write it this way. Let me write it-- Well, let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going, but I want to be fair to the colors. So c is equal to log base x of b over-- let me scroll down a little bit-- log base x, dividing both sides by that, of a. And we know from here I can just copy and paste it, this is also equal to c. This is how we defined it. So let me copy it and then let me paste it. So this is also equal to c. And we're done. We've proven the change of base formula. Log base a of b is equal to log base x of b divided by log base x of a. In this example, a was 5, b is 100, and the base that we switched it to is 10. x is 10.