If $x = 2\log_39 + \log_{27}5,$ then $3^x = ??$ If
$$x = 2\log_39 + \log_{27}5,$$
then $3^x = ??$
I know $2\log_39$ is $4$, however how do I find an exact answer to the whole equation??
I tried to change the base of the second one to $ 3 $, however it did not work out.

$$\begin{align}x= 2\log_{3}9+\log_{27}5\\
=2\log_{3}9+{1\over3}\log_{3}5 \end{align}$$
$$\begin{align}3^x=3^{[2\log_{3}9+1/3\log_{3}5]}&
&\\=3^{2\log_{3}9}3^{1/3\log_{3}5} 
&\\=9^{2\log_{3}3}5^{1/3\log_{3}3}
\\=9^{2}\cdot 5^{1/3}
\\=81\sqrt[3]5
\end{align}$$

we use :
$a^{\log(b)}=b^{\log(a)}$
proof:
Let :$ x = a^{\log(b)}$, and $y = b^{\log(a)}$ 
$\log x = \log{(a^{\log(b)})} = \log(b)\log(a) $
$\log y = \log{(b^{\log(a)})} = \log(a)\log(b) $
$\log(a)\log(b) = \log(b)\log(a)$
$ \log x = \log y$ 
$x = y $

