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

1 Answer
Jun 20, 2016

Real solution: $x = - 5$

Complex solutions: $x = - 5 + \frac{2 k \pi}{\ln} \left(2\right) i$ for any integer $k$

Explanation:

$32 = {2}^{5}$

We are given:

${2}^{x} = \frac{1}{32} = \frac{1}{{2}^{5}} = {2}^{-} 5$

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

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Complex solutions

${e}^{2 \pi i} = 1$

So:

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

So $x = - 5 + \frac{2 k \pi}{\ln} \left(2\right) i$ for any integer $k$