Powers and the logarithm By example:


*

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

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


What is the rule behind this?

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.
