Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

By example:

  • $4^{\log_2(n)}$ evaluates to $n^2$
  • $2^{\log_2(n)}$ evaluates to $n$

What is the rule behind this?

share|cite|improve this question
up vote 7 down vote accepted

The rules are:

  • $(a^b)^c = a^{bc}$
  • $a^{\log_a(b)} = b$
  • $b\log(a) = \log(a^b)$

therefore $4^{\log_2(n)} = 2^{2\log_2(n)} = 2^{\log_2(n^2)} = n^2$.

share|cite|improve this answer

$\log_ab=x$ from definition is equivalent with $a^x=b$

then replacing $x$ from first equation in second equation we get $$a^{\log_ab}=b$$ and replacing $b$ from second equation in first equation we get $$\log_aa^x=x$$

Let, $\log_ab^c=y\iff a^y=b^c=(a^x)^c=a^{xc}\implies y=xc$ if $a\ne0,1$

So, $\log_ab^c=y=cx=c\log_ab$ so

$$\log_ab^c=c\log_ab$$ then using three last equations we get $$4^{\log_2(n)}=2^{2\log_2n}=2^{\log_2n^2}=n^2$$ $$2^{\log_2n}=n$$

share|cite|improve this answer

Here's a method that relies more on applying an appropriate strategy than on formulas.

If you want to rewrite $4^{\log_2(n)}$ as a power of $n$, then you simply want to solve for $u$ in the following equation:

$$4^{\log_2(n)} = n^u$$

The standard way of solving for something that appears in the exponent of an exponentiated expression is to take the logarithm (to some fixed base) of both sides. Since $\log_2$ already shows up, we may as well take the logarithm-base-$2$ of both sides (otherwise, two different kinds of logarithms will show up, and we'd have to take care of this later): $$\log_{2}\left(4^{\log_2(n)}\right) = \log_{2}\left(n^u\right)$$ $$\log_{2}(n) \cdot \log_{2}(4) = u \cdot \log_{2}(n)$$ $$\log_{2}(4) = u$$ $$2 = u$$ In the last step I used the fact that $\log_2(4) = \log_{2}(2^2) = 2$. However, if I didn't know this, or if the numbers weren't nice (i.e. we got something like $\log_2(5) = u$), we'd still have what we wanted -- a numerical expression for the exponent of $n.$ The same method allows you to (even more easily) determine that $2^{\log_2(n)} = n.$ Here's an application of the same method to some second semester calculus (in the U.S.) semi-challenging $p$-series convergence/divergence problems.

share|cite|improve this answer

Your Answer


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.