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 $x = -5$

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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 $k$

How do you solve 2^x = 1/322^x = 1/32?