Is it possible to solve exponential equation analytically?

Question

I'm trying to solve the following equation:

$$e^{3x}-e^{2x}\left(e^2-\frac{1}{e^4}\right)-1=0$$

I know the solution is 2, as the equation above is simply a rearranged version of this initial statement:

$$e^{x}-\frac{1}{e^{2x}}=e^2-\frac{1}{e^4}$$

I assumed I could forge a cubic by letting $x=e^b$ and then using the cubic formula to do so, but I get into a hideous mess with terms being "trapped" inside cube roots and nothing really falls together nicely.

My question is, how would one go about solving this equation analytically (if it's at all possible)?

Answer 1

Because $f(x)=e^x-e^{-2x}$ increases and from here $2$ is an unique root.

There is also the following way:

We need to solve $$e^x-e^2+\frac{e^{2x}-e^4}{e^{2x+4}}=0$$ or $$(e^x-e^2)\left(1+\frac{e^x+e^2}{e^{2x+4}}\right)=0$$ or $$e^x=e^2$$ or $$x=2.$$

Answer 2

It is possible if we are clever enough to factorize out the linear term of the known root, in this case $y-e^2$. Again I pick $y=e^x$ $$y-\frac{1}{y^2} = e^2-\frac{1}{e^4}\\ \iff y-e^2 = -\left(\frac{1}{e^4}-\frac{1}{y^2}\right)\\ \iff e^4y^2(y-e^2) = -(y+e^2)(y-e^2)\\ \iff (e^4y^2+y+e^2)(y-e^2)=0\\$$ Thus our solutions are, for $e^x$, on solving the quadratic: $$e^2, \qquad \frac{-1\pm i\sqrt{4e^6-1}}{2e^4}$$ For the first, $x=2$, since in reals the exponential is increasing and for the second, note that any complex number can be expressed as $re^{i\theta}$ in which case its logarithm is $x=\ln{r} + i\theta$ which is quite messy but possible to work out if the complex numbers are of interest to you. Here we can check that that $$r=\frac{\sqrt{1+4e^6 - 1}}{2e^4} = \frac{1}{e}\qquad \theta=\pm \tan^{-1}(\sqrt{4e^6-1})\approx 1.5459 \;\left(\text{close to }88.6^{\circ}\right)$$ So we have: $\frac{1}{e} \pm i\theta$

Answer 3

Its very easy to plot $e^x-e^{-2x}$

and conclude that $e^x-e^{-2x} \in (-\infty,\infty)$

Answer 4

With $t:=e^x$ and mutliplying by $e^4$,

$$e^4t^3-t^2\left(e^6-1\right)-e^4=0$$

indeed has the solution $t=e^2$. By long division,

$$e^4t^3-t^2\left(e^6-1\right)-e^4=(t-e^2)\left(e^4t^2+ t+e^2\right)=0.$$

Now you can solve the quadratic.