# How to solve this equation?

$$\frac{4^n-1}{n-1} > 4 × 10^6$$

What is an easy way to solve it by hand? I need to find the minimum integral value of n to satisfy this equation.

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@KennyTM: thanks for the formatting. Is there some guide that can get me up to speed with math formatting? –  Lazer Aug 24 '10 at 18:07
We are using TeX markup. See meta.math.stackexchange.com/questions/480/… for detail. –  KennyTM Aug 24 '10 at 18:17

Instead of solving $\frac{4^n-1}{n-1} > 4×10^6$ try solving

$\frac{4^n}{n-1} > 4×10^6 = 4^4 ×5^6$

i.e

$\frac{4^{n-4}}{n-1} > 5^6$

Clearly $n> 10$.

Try a binary search method.

$n = 20$ is too large.

$n=15$ too large.

$n=12$ is too small.

$n=13$ seems to give the answer.

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Wolfram Alpha will solve it for you if you write: solve "your inequality". But as you want to solve it by hand you can rearrange it to: $$n > (4+3\log_2{5}) + \frac{1}{2} \log_2(n-1) - \log\left(1-\frac{1}{2^{2n}}\right)/2\log(2)$$ Now the first term in brackets is about 10.96578, use that value for n in the next log term to get 1.65849... This shows you that n>12.6, and so you get n=13. Note that the last term is too small to worry about.