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I am trying to solve the inequality $2^n <2^{2n}-1000$ using purely elementary properties of the logarithm to obtain an exact solution.

With some re-arranging and then taking the logarithm base $2$ of both sides I get:

$n+\log_2(2^n-1)>\log_2(1000)$ and I am not sure how to proceed. Any hints appreciated.

S.C.B.
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IntegrateThis
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1 Answers1

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As pointed out in the comments, let $2^n=x$. Note $$x^2-x-1000>0, x>0 \implies x>\frac{1}{2}(1+\sqrt{4001})$$

Using the quadratic formula.

So the answer is $$n> \log_{2}(1+\sqrt{4001})-1$$

S.C.B.
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  • This is brilliant, but I feel as if I would have never come up with this on my own. Do you have any quick insights into how you knew this would be a good substitution? – IntegrateThis Jan 16 '17 at 03:42
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    @JamesDickens This is a well-known substitution. I suggest you memorize it. However, the basic idea is that the term $$2^{2n}-2^{n}$$ is too difficult to handle without using substitution. – S.C.B. Jan 16 '17 at 03:44