How to solve this for $k_i$?

Question

Let $\epsilon >0$ and $$ \max_{0\leqslant i\leqslant n-1}2^{-k_i}\leqslant\epsilon. $$

How can I solve this for $k_i$?

It's the max that confuses me.

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$2^{-k_i}\leqslant\epsilon\iff k_i > \frac{\ln(1/\epsilon)}{\ln 2}$

Can I now simply take the max on both sides, getting $$ \max_{0\leqslant i\leqslant n-1}k_i > \frac{\ln(1/\epsilon)}{\ln 2}? $$

Answer 1

Hint. Note that the function $x \mapsto 2^{-x}$ is monotonically decreasing. Hence we have $$ \max_i 2^{-k_i} = 2^{-\min_i k_i}. $$ Now continue as you did.