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

1 Answer
Jun 20, 2016

Answer:

Real solution: #x = -5#

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

Explanation:

#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#