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