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Numerade Educator

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Problem 41 Hard Difficulty

Express the given quantity as a single logarithm.

$ \frac{1}{3} \ln (x + 2)^3 + \frac{1}{2} [\ln x - \ln (x^2 + 3x + 2)^2] $

Answer

$\ln \frac{\sqrt{x}}{x+1}$

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Video Transcript

Okay, We're going to simplify this expression until we get it down to the point where it is a single logger them And the first thing I would like to do is distribute the 1/2. So we have 1/3 natural log of X plus two cubed plus 1/2 natural log of X, minus 1/2 natural log of X squared plus three X plus two quantity squared. Okay, Now we're going to use the power property of lager. Them's which is going to allow us to bring these numbers out front and bring them up and make them powers inside the longer them. So for the 1st 1 it's going to be the natural log of X plus two to the third power to the 1/3 power. And if you have something to the third power and to the 1/3 power, you multiply those. And it's just to the first power for the next one. We're bringing up the 1/2 and so we're going to have X to the 1/2 power we can write extra the 1/2 power as square root X. And for the 3rd 1 we're bringing the 1/2 up making it a power and multiplying it by the power of to that was already there when half times two is one. So we have minus the natural log of X squared plus three x plus two. Okay, so now what we want to do is use other properties of logarithms and we have the product property, which tells us that if we have a log plus a log, we can combine those into the log of the product. So that would be the natural log of the product X plus two square root X. Okay. And we still have minus the other longer them. Next we're going to use the quotient property that tells us if we have a log minus a log, we can change it into the log of the quotient. So we're going to get the natural log of the quotient X plus two times the square root of X over X squared plus three X plus two. Now, you may have noticed that you actually can factor that denominator. You can factor X squared plus three x plus two. Let's do that and it factors into X plus two times X plus one So Let's go ahead and cancel the X plus twos. So our final answer as a single logger them is the natural log of the square root of X over X plus one.