# Expand $\ln\left[\frac{(4x^5-x-1)\sqrt{x-7}}{(x^2+1)^3}\right]$.

Expand this expression to the greatest possible terms with the lowest possible exponents.
$\ln\left[\dfrac{(4x^5-x-1)\sqrt{x-7}}{(x^2+1)^3}\right]$

There are two ways at which I approached this problem...
So for the first one, I started out by giving each set of parenthesis their own $\ln$ function:
$\ln(4x^5-x-1)+\ln(\sqrt{x-7})-\ln(x^2+1)^3$

My second approach was to factor out the bottom and then hopefully divide it by the top...
$\ln\left[\dfrac{(4x^5-x-1)\sqrt{x-7}}{x^6+3x^4+3x^2+1}\right]$
And my next plan was to divide $4x^4-x-1$ by $x^6+3x^4+3x^2+1$

Can someone tell me which approach is the correct way, or if they are both wrong. Please do not give full answers' only hints.

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You need to check your denominator and what you are doing with that term. As you have written the question you have given it three different values. –  Mark Bennet Aug 21 '12 at 8:02
What do you mean? Doesn't $(x^2+1)(x^2+1)(x^2+1)=(x^8-x^2-1)$? OOps, I had a sign mix-up on my paper. Fixing now. –  Austin Broussard Aug 21 '12 at 8:03
@AustinBroussard In the first expression you wrote $4x^{\color{red}{5}}$, and in the next big one, $4x^{\color{red}{4}}$. –  FrenzY DT. Aug 21 '12 at 8:10
You might be expected to use $\log r^t = t \log r$ –  Mark Bennet Aug 21 '12 at 8:30
Are you after an expansion when $x\to+\infty$? –  Did Aug 21 '12 at 9:38

I think what you're expected to do is this. Start with your first approach: $$\log(4x^5-x-1)+\log\sqrt{x-7}-\log\left((x^2+1)^3\right)$$ Now: do you know how to write $\sqrt{x-7}$ in the form $(x-7)^q$ for some cleverly chosen $q$? and do you know how to write $\log a^b$ without any exponents in it? If you can do those two things, you can get the answer that I expect is the intended answer. Note in particular that you can't do anything more with the $\log(4x^5-x-1)$; that already has as many terms with as low exponents as possible.

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+1 for managing to give hints without flatly writing down the answer. –  Did Aug 21 '12 at 12:43
$\log(4x^5-x-1)+\frac{1}{2}\log(x-7)-3\log(x^2+1)$? –  Austin Broussard Aug 21 '12 at 18:04
Yes.${}{}{}{}{}$ –  Gerry Myerson Aug 22 '12 at 2:18

You could try using: $\ln x^d = d \ln x$ in the following way:

$$\ln\left[\dfrac{(4x^5-x-1)\sqrt{x-7}}{(x^2+1)^3}\right]$$ $$= \ln\left[\dfrac{(4x^5-x-1)\sqrt{x-7}}{(x^2+1)^3}\right]^{(2)(\frac{1}{2})}$$ $$= \frac{1}{2} \ln\left[\dfrac{(4x^5-x-1)\sqrt{x-7}}{(x^2+1)^3}\right]^2$$ $$= \frac{1}{2} \ln\left[\dfrac{(4x^5-x-1)^2(x-7)}{(x^2+1)^6}\right]$$

At least you'll get rid of that horrible square root!

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