How to solve an exponential and logarithmic system of equations? $$
\left\{\begin{array}{c}
e^{2x} + e^y = 800 
\\ 3\ln(x) + \ln(y) = 5 
\end{array}\right.$$
I understand how to solve system of equations, logarithmic rules, and the fact that $\ln(e^x) = e^{\ln(x)} = x$. However, any direction I seem to go with this problem causes me difficulties. For example, when solving for y, I can get $y = \ln\left(800 - e^{2x}\right)$ or $y = \frac{e^5}{x^3}$. Both directions seem to lead me into a dead end. Although system of equations seems to be a frequent question on here, I haven't found anything helpful towards this problem.

HINT:
We have $$e^{2x}+e^{y}=800\tag 1$$ $$3\ln(x)+\ln(y)=5\tag 2$$ 
$$\implies \ln(x^3)+\ln(y)=5$$ 
$$\ln(x^3y)=5$$ $$x^3y=e^5$$ $$y=\frac{e^5}{x^3}$$ setting this value of $y$ in (1), we get 
$$e^{2x}+e^{(e^5/x^3)}=800$$
$$e^{2x}+e^{(e^5/x^3)}-800=0$$
Can you proceed further?
