$$ \left\{\begin{array}{c} e^{2x} + e^y = 800 \\ 3\ln(x) + \ln(y) = 5 \end{array}\right.$$

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.


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).

  • $\begingroup$ I want to add, that although we hoped for only one solution I found another one between x=3.3 and x=3.4...so another try with Newton and starting point 3.3 could be very promising $\endgroup$ – user190080 Aug 23 '15 at 15:37
  • $\begingroup$ Great ! Please, post an answer with that ! This equation is just a monster ! Cheers ;-) $\endgroup$ – Claude Leibovici Aug 23 '15 at 16:10
  • $\begingroup$ Oh sorry for the misunderstanding, I actually hoped you could add this to your answer! I was just playing around and was using my computer merely as a big calculator, I got using Newton $x=3.403$, I haven't found a closed form solution. $\endgroup$ – user190080 Aug 23 '15 at 16:44
  • $\begingroup$ @user190080. I hope that closed form is the joke of the year !! No, please, put an answer giving the iterates. I will be intesting to compare the path to the second root (as you probably noticed, I had an overshoot of the solution). Again, do it ! $\endgroup$ – Claude Leibovici Aug 23 '15 at 17:26
  • $\begingroup$ @user190080. By the way, I started with computers 56 years ago and, for me, these are only big and dumb calculators. $\endgroup$ – Claude Leibovici Aug 23 '15 at 17:29

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

  • 1
    $\begingroup$ It's fairly easy to show that there's only one possible solution to $f'(x) = 0$ (since $x^4 f'(x)$ is strictly increasing by inspection) for positive $x$, and hence at most two solutions to $f(x) = 0$ for positive $x$ (and clearly, no solutions for negative $x$). $\endgroup$ – Mark Dickinson Aug 23 '15 at 18:49
  • $\begingroup$ @MarkDickinson Nice! I agree, it is indeed not hard to see, I just haven't checked it... $\endgroup$ – user190080 Aug 23 '15 at 19:03


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?

  • 2
    $\begingroup$ No, that's actually where I'm stuck. I already solved for y and substituted it into the other equation. $\endgroup$ – Ren Aug 23 '15 at 14:49
  • $\begingroup$ Q: "Can you proceed further ?" A: "No, but I'm interested in how it works." $\endgroup$ – callculus Aug 23 '15 at 15:01
  • 3
    $\begingroup$ This "answer" says nothing OP did not already know. $\endgroup$ – tomasz Aug 23 '15 at 19:07
  • $\begingroup$ Could some one tell me the reason what is wrong with this hint? $\endgroup$ – Harish Chandra Rajpoot Aug 24 '15 at 2:47

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.


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

  • $\begingroup$ You may have noticed that you did not write down a system of two linear equations, and I suspect that this approach will not succeed. However if you can give such a solution, I will gladly up vote. $\endgroup$ – hardmath Aug 24 '15 at 3:10

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