Hi!

Solve $\displaystyle e^{2x} = 3^{x - 4}$

I have used $\displaystyle \log_3$ but this is where it leads:

$\displaystyle \log_3 e^{2x} = \log_3 3^{x-4}$

$\displaystyle 2x \log_3 e = x - 4$

$\displaystyle 2x \log_3 e + 4 = x$

$\displaystyle \frac{\log_3 e + 4}{2} = \frac{x}{2x} = \frac12$