Question

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

Answer 1

There are no Real solutions, but:

`x = (+-pi/3+2kpi)i` for any integer `k`

Explanation:

`color(white)()`
Method 1 - trigonometric

Let `x = it` and divide both sides of the equation by `2` to get:

`1/2 = (e^(it)+e^(-it)) / 2 = cos(t)`

So `t = +-cos^(-1)(1/2) + 2kpi = +-pi/3 + 2kpi`

So `x = t/i = (+-pi/3+2kpi)i`

`color(white)()`
Method 2 - logarithm

Let `t = e^x`

Then this equation becomes:

`t+1/t = 1`

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

`0 = t^2-t+1`

`= (t-1/2)^2+3/4`

`= (t-1/2)^2-(sqrt(3)/2i)^2`

`= (t-1/2-sqrt(3)/2i)(t-1/2+sqrt(3)/2i)`

So:

`e^x = t = 1/2+-sqrt(3)/2i`

Hence:

`x = ln(1/2+-sqrt(3)/2i) + 2kpii` for any integer `k`

Note that we can add any integer multiple of `2pii` since `e^(2pii) = e^(-2pii) = 1`

Now:

`ln(1/2+-sqrt(3)/2i) = ln abs(1/2+-sqrt(3)/2i) + Arg(1/2+-sqrt(3)/2i) i`

`=ln (sqrt((1/2)^2+(sqrt(3)/2)^2)) +- tan^(-1)(sqrt(3))`

`= ln (1) +- pi/3`

`= +-pi/3`