How do you solve #2^x = 1/32#?

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Answer 1 · 2016-06-20T21:43:49

Real solution: #x = -5#

Complex solutions: #x = -5 + (2kpi)/ln(2) i# for any integer #k#

#32 = 2^5#

We are given:

#2^x = 1/32 = 1/(2^5) = 2^-5#

So the unique Real solution is #x = -5#

#color(white)()#
Complex solutions

#e^(2pii) = 1#

So:

#2^x = 2^-5*e^(2kpii) = 2^(-5)*2^((2kpi)/ln(2) i) = 2^(-5+(2kpi)/ln(2)i#

So #x = -5 +(2kpi)/ln(2) i# for any integer #k#