The goal of the problem is to solve for x using logs and the given equation is this:
2e^(3x) = 4e^(5x)
Any help on how to solve this?
I'll start you off;
$\displaystyle 2e^{3x}=4e^{5x}$
$\displaystyle e^{3x}=2e^{5x}$
$\displaystyle ln(e^{3x})=ln(2e^{5x})$
$\displaystyle 3x=ln(2)+ln(e^{5x})$
Can you take it from here?
Keep in mind the handy rules $\displaystyle log_{a}a=1$, and $\displaystyle log_{a}b^c=c \cdot log_{a}b$