Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

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 :)

share|improve this question
1  
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
1  
I edited to remove the tag "calculus", because I feel "algebra-precalculus" is where this belongs. – Alex Becker Dec 28 '11 at 19:24

1 Answer

up vote 4 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.