How do you solve e^(x) + e^(-x) = 1?

Jun 13, 2016

There are no Real solutions, but:

$x = \left(\pm \frac{\pi}{3} + 2 k \pi\right) i$ for any integer $k$

Explanation:

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Method 1 - trigonometric

Let $x = i t$ and divide both sides of the equation by $2$ to get:

$\frac{1}{2} = \frac{{e}^{i t} + {e}^{- i t}}{2} = \cos \left(t\right)$

So $t = \pm {\cos}^{- 1} \left(\frac{1}{2}\right) + 2 k \pi = \pm \frac{\pi}{3} + 2 k \pi$

So $x = \frac{t}{i} = \left(\pm \frac{\pi}{3} + 2 k \pi\right) i$

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Method 2 - logarithm

Let $t = {e}^{x}$

Then this equation becomes:

$t + \frac{1}{t} = 1$

Multiply through by $t$ and rearrange a little to get:

$0 = {t}^{2} - t + 1$

$= {\left(t - \frac{1}{2}\right)}^{2} + \frac{3}{4}$

$= {\left(t - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{3}}{2} i\right)}^{2}$

$= \left(t - \frac{1}{2} - \frac{\sqrt{3}}{2} i\right) \left(t - \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

So:

${e}^{x} = t = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

Hence:

$x = \ln \left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\right) + 2 k \pi i$ for any integer $k$

Note that we can add any integer multiple of $2 \pi i$ since ${e}^{2 \pi i} = {e}^{- 2 \pi i} = 1$

Now:

$\ln \left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\right) = \ln \left\mid \frac{1}{2} \pm \frac{\sqrt{3}}{2} i \right\mid + A r g \left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\right) i$

$= \ln \left(\sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}}\right) \pm {\tan}^{- 1} \left(\sqrt{3}\right)$

$= \ln \left(1\right) \pm \frac{\pi}{3}$

$= \pm \frac{\pi}{3}$