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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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    I edited to remove the tag "calculus", because I feel "algebra-precalculus" is where this belongs.2011-12-28

1 Answers 1

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