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.
 A: So here is another solution for your problem, again using Newton's method we get by using as initial value $x_0=3.2$ (it is very sensitive!) the following line of iterates
$$x_0=3.2$$ 
$$x_1=3.330187$$ 
$$x_2=3.303383$$ 
$$x_3=3.302208$$ 
$$x_4=3.302206$$
which gives us as an approximate solution
$$
f(3.302206)\approx-5.277731 \cdot10^{-5}
$$
Of course it still possible that there are other solutions outside just waiting for us to be discovered, a graph analysis might give some clearance about that. 
For the graph analysis please check the comment by Mark just below the answer! tl;dr: there are at most two solutions + there are at least two solutions $\to$ there are exactly two solutions 
A: 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?
A: 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 $a < 632.498$. 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).
A: Your system of equations are transcendental, numerical methods work better. 
If $ f(x)= e^{e^x}$ is a standard function, an expression for them may be 
written. 
A: 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
