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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.

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    Set $x=2^n$ and solve the resulting quadratic. – 2017-01-16
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    You forgot to take $\log_2$ of $1000$ in your calculation. Is $n$ supposed to be an integer or real number? – 2017-01-16
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    @B.Goddard a real number – 2017-01-16

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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 $$n> \log_{2}(1+\sqrt{4001})-1$$

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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? – 2017-01-16
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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. – 2017-01-16