1. ## Algebraic deduction help

Hi Guys,

I need a little help with the algebra in this inequality. I have made a bit of mess of it.

Given,

$\displaystyle \dfrac{100x^{n+1}}{1 - x^{n+1}} < \dfrac{k}{100}$

Deduce, that n must exceed,

$\displaystyle \dfrac{log \frac{100+k}{k}}{log \frac{1}{x}} - 1$

Thanks.

EDIT: Not sure why the latex code is giving an error. I checked it in MikTex and it works fine?

2. Originally Posted by mathguy80
EDIT: Not sure why the latex code is giving an error. I checked it in MikTex and it works fine?
There seem to be problems with LaTeX in MHF after yesterday's shutdown.

I got to the required conclusion under a couple of assumptions, but I have 100^2 instead of 100. First, if 1 - x^{n+1} > 0 and 100^2 + k > 0, then the initial inequality can be solved for x^{n+1} to get

x^{n+1} < k / (100^2 + k).

Then if x > 0, we have

(1 / x)^{n+1} > (100^2 + k) / k.

Taking log of both sides and dividing by log(1 / x) (which requires that log(1 / x) > 0, i.e., 1 / x > 1, i.e., x < 1, which accords with 1 - x^{n+1} > 0 above), you get the required inequality.

3. Thanks @emakarov. I think the required answer may be wrong. Your solution looks good to me. I had not make the assumption to divide by log 1/x. Good to know, thanks.