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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)?

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  • $\begingroup$ $e^{x} - e^{-2x} = e^{2} - e^{-4}$ do a little more factoring, and do a substitution but I think it's pretty miserable to try to do this analytically $\endgroup$ – user29418 Sep 24 '20 at 17:41
  • $\begingroup$ Have you tried just multiplying the whole thing with $e^4$ and then applying your idea? $\endgroup$ – Zeno Sep 24 '20 at 17:43
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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 $$x=2.$$

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    $\begingroup$ Thank you, that's a beautiful method; I faced a similar problem to the one you have answered and you have given me a new and much better way to solve it. It is much appreciated. $\endgroup$ – A-Level Student Sep 24 '20 at 20:59
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    $\begingroup$ Thanks! I didn't see this, but it totally makes sense! $\endgroup$ – Daniel Podobinski Sep 25 '20 at 17:11
  • $\begingroup$ You are welcome, @Daniel $\endgroup$ – Michael Rozenberg Sep 25 '20 at 17:28
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Its very easy to plot $e^x-e^{-2x}$

x

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

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  • $\begingroup$ That's not the equation I posted unfortunately. $\endgroup$ – Daniel Podobinski Sep 25 '20 at 15:12
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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: $\frac{1}{e} \pm i\theta$

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  • $\begingroup$ Thank you! I used another posters method but this is equally interesting. $\endgroup$ – Daniel Podobinski Sep 27 '20 at 22:50
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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.

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