Solving $2^{16^x} = 16^{2^x}$ 
Find $x$ if $2^{16^x} = 16^{2^x}$.

If $x = 0$ we have $2=16$.
$8$ times or $(2^3)$ difference and if $x = 1$ we have $65536=256$.
$256$ or $2^8$ difference. I can't see anything useful in figuring this out.
I have a feeling $x$ must be a fraction. I know only know about logarithm. I haven't learned anything about the number $e$ or $ln$ yet.

If you take the $\log_2$ of both sides you end up with
\begin{align*}
\log_2(2^{16^x}) &= 16^x \log_2(2) = 16^x = 2^x\cdot8^x\\
\log_2(16^{2^x}) &= 2^x\log_2{16} = 2^x\cdot4
\end{align*}
From this you get that $2^x\cdot 4 = 2^x \cdot 8^x \implies 8^x = 4 \implies x =\log_{8}(4) = 2/3.$
($\log_8(4) = \log_8(2^2) = 2\log_8(2) = 2\cdot1/3$)
