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

1 Answer
Jun 13, 2016

Answer:

There are no Real solutions, but:

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

Explanation:

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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#

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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#