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Let's see if we can stumble our way to another

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logarithm property.

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So let's say that the log base x of A is equal to B.

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That's the same thing as saying that x to the B is equal to A.

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Fair enough.

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So what I want to do is experiment.

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What happens if I multiply this expression by another variable?

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Let's call it C.

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So I'm going to multiply both sides of this equation times C.

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And I'll just switch colors just to keep things

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interesting.

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That's not an x that's a C.

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I should probably just do a dot instead.

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Times C.

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So I'm going to multiply both sides of this equation times C.

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So I get C times log base x of A is equal to-- multiply both

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sides of the equation-- is equal to B times C.

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Fair enough.

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I think you realize I have not done anything

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profound just yet.

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But let's go back.

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We said that this is the same thing as this.

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So let's experiment with something.

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Let's raise this side to the power of C.

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So I'm going to raise this side to the power of C.

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That's a kind of caret.

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And when you type exponents that's what

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you would use, a caret.

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So I'm going to raise it to the power of C.

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So then, this side is x to the B to the C power,

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is equal to A to the C.

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All I did is I raised both sides of this equation

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to the Cth power.

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And what do we know about when you raise something to an

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exponent and you raise that whole thing to another

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exponent, what happens to the exponents?

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Well, that's just exponent rule and you just multiply

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those two exponents.

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This just implies that x to the BC is equal to A to the C.

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What can we do now?

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Well, I don't know.

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Let's take the logarithm of both sides.

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Or let's just write this-- let's not take the

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logarithm of both sides.

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Let's write this as a logarithm expression.

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We know that x to the BC is equal to A to the C.

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Well, that's the exact same thing as saying that the

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logarithm base x of A to the C is equal to BC.

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Correct?

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Because all I did is I rewrote this as a logarithm expression.

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And I think now you realized that something

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interesting has happened.

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That BC, well, of course, it's the same thing as this BC.

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So this expression must be equal to this expression.

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And I think we have another logarithm property.

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That if I have some kind of a coefficient in front of the

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logarithm where I'm multiplying the logarithm, so if I have C

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log-- Clog base x of A, but that's C times the

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logarithm base x of A.

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That equals the log base x of A to the C.

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So you could take this coefficient and instead make

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it an exponent on the term inside the logarithm.

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That is another logarithm property.

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So let's review what we know so far about logarithms.

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We know that if I write-- let me say-- well, let me just with

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the letters I've been using.

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C times logarithm base x of A is equal to logarithm

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base x of A to the C.

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We know that.

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And we know-- we just learned that logarithm base x of A plus

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logarithm base x of B is equal to the logarithm base

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x of A times B.

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Now let me ask you a question.

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What happens if instead of a positive sign here we

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put a negative sign?

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Well, you could probably figure it out yourself but we could do

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that same exact proof that we did in the beginning.

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But in this time we will set it up with a negative.

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Let's just say that log base x of A is equal to l.

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Let's say that log base x of B is equal to m.

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Let's say that log base x of A divided by B is equal to n.

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How can we write all of these expressions as exponents?

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Well, this just says that x to the l is equal to A.

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Let me switch colors.

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That keeps it interesting.

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This is just saying that x to the m is equal to B.

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And this is just saying that x to the n is equal to A/B.

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So what can we do here?

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Well what's another way of writing A/B?

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Well, that's just the same thing as writing x to the l

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because that's A, over x to the m.

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That's B.

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And this we know from our exponent rules-- this could

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also be written as x to the l, x to the negative m.

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Or that also equals x to the l minus m.

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So what do we know?

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We know that x to the n is equal to x to the l minus m.

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Those equal each other.

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I just made a big equal chain here.

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So we know that n is equal to l minus m.

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Well, what does that do for us?

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Well, what's another way of writing n?

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I'm going to do it up here because I think we have

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stumbled upon another logarithm rule.

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What's another way of writing n?

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Well, I did it right here.

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This is another way of writing n.

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So logarithm base x of A/B-- this is an x over

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here-- is equal to l.

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l is this right here.

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Log base x of A is equal to l.

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The log base x of A minus m.

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I wrote m right here.

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That's log base x of B.

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There you go.

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I probably didn't have to prove it.

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You could've probably tried it out with dividing

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it, but whatever.

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But you know are hopefully satisfied that we have this new

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logarithm property right there.

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Now I have one more logarithm property to show you, but I

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don't think I have time to show it in this video.

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So I will do it in the next video.

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I'll see you soon.

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