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$$ \frac{4^n-1}{n-1} > 4 × 10^6 $$

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
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3 Answers

up vote 2 down vote accepted

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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If n has to be an integer, you can find the solution by brute force. Once you have the solution, you can show that it is correct and unique by showing that the function f(n) = (4^n - 1)/(n-1) is increasing for n >= 2.

If n is allowed to be real, you can use your integer guess as the starting point for Newton's method and find the solution numerically to as much precision as you'd like. Since f(n) is increasing, you can even find a good interval estimate. if f(a) < 4*10^6 and f(b) > 4*10^6, you know the solution is in the interval (a, b).

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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.

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I'm not sure how to compute the logarithm easily by hand... –  KennyTM Aug 28 '10 at 19:37
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