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Let’s see if we can stumble our way to another logarithm property. So, let’s say that the log base x of A is equal to B. That is exact same thing as saying that X to the B is equal to A, fair enough.
So, what I want to do is experiment. What happens if I multiply this expression by another variable, let’s call it C. So, I’m going to multiply both sides of this equation times C. So, I’m going to multiply both sides of this equation times C. So, I get C times log base X of A is equal to—and multiply both sides of equation. It’s equal to B times C, fair enough. I think you realize I have not done anything profound just yet.
Well, let’s go back. We said that this is the same thing as this so let’s experiment with some thing. Let’s raise this side to the power C. So, I’m going to raise this side to the power C, that’s kind of a current. And when you type exponents, that’s what you do with the current, so I’m going to raise it to the power C. So then, this sides is X to the B to the C power is equal to A to the C. All you do this is to raise both sides of this equation to the C power.
And what do we know about when you raise something to an exponent and you raise that whole thing to another exponent, what happens to the exponents? Well, that’s just an exponent rule and you just multiply those two exponents. So, this just implies that X to the BC is equal to A to the C. What can we do now? Let’s take the logarithm of both sides or let’s just write this. Let’s not take the logarithm of both sides, let’s write this as a logarithm expression. We know that X to the B C is equal to A to the Z. Well, that’s the exact same thing as saying that the logarithm base X of A to the C, logarithm base X of A to Z is equal to BC, correct? Because all I did is I rewrote this as a logarithm expression.
And I think now you realize that something interesting has happened. That BC, this BC, well of course it’s the same thing as this BC. So, this expression must be equal to this expression and I think we have another logarithm property. That if I have some kind of coefficient in front of the logarithm, we’re multiplying the logarithm so I have C log base X of A but that C times the logarithm base X of A, that equals the log base X of A to the C. So, you could take this coefficient and instead make it an exponent on the turning side of the logarithm. That is another logarithm properties so let’s review what we know so far about logarithms. We know that if I write—let me say C times logarithm base X of A is equal to logarithm base X of A to the C. We know that and we know, we just learned that logarithm base X of A plus logarithm base X of B is equal to the logarithm base X of A times B.
Now, let me ask you a question. What happens if instead of a positive sign here, we put a negative sign? Well, you could probably figure out your self but we could do that same exact proof that we did in the beginning but in this time we will set it up with a negative. So, if I said that log base X of A is equal to L, let’s say that log base X of B is equal to M, let’s say that log base X of A divided by B is equal to N. How can we write all of these expressions and exponents? Well, this just says that X to the L is equal to A. This is just saying that X to the M is equal to B and this is just saying that X to the N is equal to A over B.
So, what can we do here? What is another way of writing A/B? Well, that’s just the same thing as writing X to the L because that’s A, X to the L over X to the M, that’s B. And this we know from our exponent rules, you could write and this could also be written as X to the L, X to the negative M or that also equals X to the L minus M.
So, what do we know? We know that X to the N is equal to X to the L minus M. X to the N is equal to X to the L minus M. Those equal to each other, I just made a big equal chain here. So, we know that N is equal to L minus M. But what is that do for us? What’s another way of writing N? I’m going to do it up here because I think we have stumbled upon another logarithm rule.
What’s another way of writing N? I did it right here. This is another way of writing N. So logarithm, logarithm base X of A over B is equal to L. L is this right here. Log base X of A is equal to L. The log base X of A minus M. That’s log base X of B, there you go. I probably did not have to prove. You could have probably tried it out with the dividing of—whatever but you now are hopefully satisfied that we have this new logarithm property right there.
Now, I’ve one more logarithm property to show you but I don’t think I have time to show in this video so I will do it in the next video. I’ll see you soon.
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