# Solve for x to three significant digits?

## ${e}^{x} + {e}^{-} x = 2$

Dec 13, 2016

Real solution: $x = 0$

Complex solutions: $x = 2 k \pi i$ for any integer $k$

#### Explanation:

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

Then our equation becomes:

$t + \frac{1}{t} = 2$

Mutliply through by $t$ and subtract $2 t$ from both sides to find:

$0 = {t}^{2} - 2 t + 1 = {\left(t - 1\right)}^{2}$

So the only possible value of $t$ is $t = 1$

Now solve ${e}^{x} = 1$

The unique Real solution is $x = 0$ since ${a}^{0} = 1$ for all $a \ne 0$

There are also Complex solutions resulting from Euler's identity:

${e}^{i \pi} = - 1$

Hence:

${e}^{\left(2 k \pi\right) i} = {\left({\left({e}^{i \pi}\right)}^{2}\right)}^{k} = {\left({\left(- 1\right)}^{2}\right)}^{k} = {1}^{k} = 1$

for any integer $k$