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

1 Answer
George C.
Jun 13, 2016

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

Related questions
See all questions in Exponential and Logistic Functions on a Graphing Calculator
Impact of this question
4389 views around the world
You can reuse this answer
Creative Commons License