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.

Use the fact that $\log_{27}(5) = \frac{\log_3(5)}{\log_3(27)} = \frac{\log_3(5)}{3} = \log_3(5^{\frac{1}{3}})$.
Then $x = 2\log_3(9) + \log_{27}(5)  = \log_3(81) + \log_3(5^\frac{1}{3}) = \log_3(81 \cdot 5^{\frac{1}{3}})$.
You obtain that $3^x = 81 \cdot 5^\frac{1}{3}$.
