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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.2012-08-21
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    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.2012-08-21
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    @AustinBroussard In the first expression you wrote $4x^{\color{red}{5}}$, and in the next big one, $4x^{\color{red}{4}}$.2012-08-21
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    Typo... thanks for that2012-08-21
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    $(x^2+1)(x^2+1)(x^2+1)=(x^{\color{red}{6}} + 3x^4 + 3x^2 + 1)$2012-08-21
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    @Arthur i was thinking of the wrong exponential rule..2012-08-21
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    You have also got $x^2+1$ in your first formula, but $x^2-1$ in your first approach2012-08-21
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    You might be expected to use $\log r^t = t \log r$2012-08-21
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    I'm glad you guys can read, I'm just so tired and I need to sleep after this problem. I can't sleep with an unknown answer!2012-08-21
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    I know that, but which approach should I use?2012-08-21
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    Which has the lower exponent $x^6$ or $x^2$? There is more problem to know what to do with the square root term, because the question is rather unclearly phrased.2012-08-21
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    Well, the square root has the lowest exponent of $\frac{1}{2}$.2012-08-21
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    How would I write $\ln(4x^5-x-1)$ in $\log r^t=t\log r$ form?2012-08-21
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    Are you after an expansion when $x\to+\infty$?2012-08-21

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