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I know it's a easy question but unfortunately I forgot some school stuff:

I have $k=\log_2(N)$ and want to know $N$.

Is it $N=2^k$ while using $2$ as base?

Short comments are welcome :)

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Yes, that's correct. – Alex Becker Dec 28 '11 at 19:23
Change it to an answer, so I can check it :) – ulead86 Dec 28 '11 at 19:24
I edited to remove the tag "calculus", because I feel "algebra-precalculus" is where this belongs. – Alex Becker Dec 28 '11 at 19:24
up vote 4 down vote accepted

Yes, that's correct. More generally, if $k = \log_b(N)$ then $N = b^{\log_b(N)} = b^k$.

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