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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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    Are you after an expansion when $x\to+\infty$?2012-08-21

1 Answers 1

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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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    Yes.${}{}{}{}{}$2012-08-22