How do you solve #e^(x) + e^(-x) = 1#?
1 Answer
There are no Real solutions, but:
#x = (+-pi/3+2kpi)i# for any integer#k#
Explanation:
Method 1 - trigonometric
Let
#1/2 = (e^(it)+e^(-it)) / 2 = cos(t)#
So
So
Method 2 - logarithm
Let
Then this equation becomes:
#t+1/t = 1#
Multiply through by
#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
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#