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Combine the given expression into a single logarithm.$$3\left[\log _{3}(x+1)-2 \log _{3}(x-1)\right]-2\left[\log _{3}(x-1)+\log _{3}(x+1)\right]$$

$$\log _{3} \frac{x+1}{(x-1)^{8}}$$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 6

Properties of Logarithmic Functions

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

00:21

Rewrite each expression as…

01:21

Write each expression as a…

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01:16

Simplify to a single logar…

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Condense the expression to…

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Combine the given expressi…

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Write each logarithmic exp…

eso. Here's another problem that has a lot of different ways of getting the correct answer. Eso I'm just sort of I mean, I show you one way and hopefully it makes sense on if it doesn't, Then I would say, you know, feel free to try to work it out a different way on, see if you get the same answer. And you should, because there's only one correct simplification. Eso Right now I'm just copying down the problem. Log based three, uh, X minus one. I'm just using on wild logs and algebra manipulation. Oh, so what I would do first is just use algebra and distribute these three and this negative to into the problem. So what I'm looking at is three log based three of X plus one. I've been changed colors here just to emphasize that that's minus six Log based three of X minus one and then distributing negative to in their native to log based three of X minus one sometimes forgot to write to. And they have two types. Uh, well, one being negative to log based three of X plus one and I try to color code this so you can see the same terms. We have a log based three of X plus one and then a later, minus two log based three of X plus one so it can combine them together to be one log based three of X plus one. And then the other piece is negative six minus two. So I'd be negative. Eight log based, three of X minus one. So then, from here, uh, the the numbers that air in front can be written as exponents. Um, but this number in front, it doesn't really matter, because something to the first power is the same. Thing is, you know, not having anything. Um And so the other piece that we have to combine is the subtraction. Well, we can combine logs together as a single law algorithm with subtraction by using division. So the this X plus one goes into the numerator. We don't have to have parentheses, but some people like to have it. And then this x minus one goes into the denominator because of the minus. And don't forget about that new exponents to the eighth power on. This is your final answer. And if I was better at math, I would be writing equal signs all the way through. I just usually forget to write them. Anyway, this is your correct answer.

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