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We're asked to simplify log base 5 of x to the third

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and once again we're just going to rewrite this in a different way

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you could argue whether's going to be more simple or not

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and the logarithm property that I'm guessing we should use for

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this example right here is

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the property if I take log base x

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of, of, let me fix more letters here

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of log base x of y to the zth power

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that this is the same thing as z times

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log base x of y

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so this is the logarithm property. If I'm taking the logarithm of a given base

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of something to the power, I could take that power upfront

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and multiply that times the log of the base of

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just the y in this case. So we apply this property over here

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and the second once I do this problem I'll talk about why this actually makes a lot of sense

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it comes straight out of exponent properties. If we just apply

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that over here we get log base 5 of x to the

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third. Well this is the exponent right over here. It's the same thing

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as z so that's going to be the same this as

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that three is the same thing. We could put it out front

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that's the same thing as three times the logarithm

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base five of x. And we're done this is just

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another way of writing it using this property so you can argue this is a

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maybe this is a simplification because you took the exponent outside of the logarithm and you're multiplying

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the logarithm by that number now. Now that out of the way, lets think about why that actually makes sense.

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So lets say that we know that "a" to the "b" power is equal to "c"

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and so if we know that, that's written as an exponential equation, if we want to write

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the same truth as a logarithmic equation

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we would say, "logarithm base 'a' of 'c' is equal to 'b'"

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To what power do I have to raise 'a' to get 'c'?

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I raise it to the bth power

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'a' to the 'b' power is equal to 'c'. Fair enough.

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So lets take both sides of the equation over here and raise it to the 'dth' power.

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Instead of doing in place I'm just going to rewrite it over here.

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So I wrote the original equation, 'a' to the 'b' is equal to 'c', which is just rewriting the statement.

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So let me just take both sides of this to the dth power and I should be consistent

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I'll use all capital letters so this should be a 'b'

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actually I'm using all lower cases so this is a lower case 'c'

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I'm going to raise this to the dth power and I'm going to raise this to the dth power.

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Obviously these two are equal to each other if I raise both sides to the same power

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The equality is still going to hold.

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Now, what's over here

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is we can use what we know about exponent properties

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Say if I have 'a' to the bth power, then I raise that to the dth power

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our exponent properties say that this is the same thing

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this is equal to 'a' to the b*d power

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this right over here using what we know about exponent properties

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this is equal to 'a' to the b<i>d power. So we have 'a' to the b</i>d power

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so we have a to the b<i>d power this is equal to c to the b</i>d power

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Now this exponential equation, if we write it as a logarithmic equation

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we would say, log base 'a' of 'c' to the dth power

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is equal to b*d. What power do I have to raise 'a' to

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to get to 'c' to the dth power? To get to this I have to raise it to the b*d power.

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But what do we know that 'b' is? We already know that 'b' is this thing right over here.

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So if we substitute this in for 'b' and we can write this as b*d

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we get logarithm base 'a' c to the dth power is equal to b<i>d or you can also call that d</i>b

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and so that's equal to d times b, be is just

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log base 'a' of 'c'

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So there you have it we just derived this property

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log base 'a' of c to the dth, that's the same thing as d times log base 'a' of c

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which we applied right over here.