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.
Having  set your equation as Harish Chandra Rajpoot answered $$f(x)=e^{2x}+e^{(e^5/x^3)}-800$$ each piece varies very fast. So, hoping that only one solution exist, try inspection $$f(1)\approx 2.85112\times 10^{64}$$ $$f(2)\approx 1.13992\times 10^8$$ $$f(3)\approx-152.665$$ So, the solution is very close to $3$. Since we have a "reasonable" guess, let us use Newton method which, starting at the guess $x_0=3$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For your case $$f'(x)=2 e^{2 x}-\frac{3 e^{\frac{e^5}{x^3}+5}}{x^4}$$ 

Applying the method, the successive iterates are $$x_1=2.71403$$ $$x_2=2.79701$$ $$x_3=2.85671$$ $$x_4=2.88250$$ $$x_5=2.88649$$ $$x_6=2.88657$$  which is the solution for six significant figures.

Edit

As Mark Dickinson commented, the first derivative of the function cancels only once for $x_*\approx 3.059$ and for this value $e^{2x_*}+e^{(e^5/x_*^3)}\approx 632.498$ which means that equation $e^{2x}+e^{(e^5/x^3)}=a$ would not show any solution if $a &lt; 632.498$. Building a Taylor expansion around $x=3$, we have $$f(x)=\left(-800+e^6+e^{\frac{e^5}{27}}\right)-\frac{1}{27} \left(e^5
   \left(e^{\frac{e^5}{27}}-54 e\right)\right) (x-3)+\frac{e^5 \left(2916 e+36
   e^{\frac{e^5}{27}}+e^{5+\frac{e^5}{27}}\right)
   }{1458}(x-3)^2+O\left((x-3)^3\right)$$ the roots of which being $\approx 2.62606$ and $\approx 3.47307$ which are very good guesses of the solution given above and of the extra solution identified by @user190080.

It would have been a better idea to look at the roots of function $$g(x)=\log\Big(e^{2x}+e^{(e^5/x^3)}\Big)-\log(800)$$ which is much better conditionned (at least from a numerical point of view).
You have a set of 2 logarithmic equations, but using the rules of logarithms, you can turn them into a set of simplier equations.

$ln(e^{2x} + e^{y}) = ln(800)$ for the first equation and $e^{3ln(x) + ln(y)} = e^{5}$ for the second.

Solving using your logarithmic rules the gives you a set of two equations which you can solve any way you choose.

New system, $2x + y = ln(800)$ and $x^3 + y = e^5$, gives you a real solution at $x \approx  5.34162$ and$ y \approx -3.99862$ along with two complex solutions.

edit: my mistake, not a set of linear equations