How do you solve $2^{x} = \frac{1}{32}$ $2^x = 1/32$ ?

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How do you solve $2^{x} = \frac{1}{32}$ $2^x = 1/32$ ?

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Answer 1 · 2016-06-20T21:43:49

Real solution: $x = - 5$ $x = -5$

Complex solutions: $x = - 5 + \frac{2 k π}{ln (2)} i$ $x = -5 + (2kpi)/ln(2) i$ for any integer $k$ $k$

Explanation:

$32 = 2^{5}$ $32 = 2^5$

We are given:

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

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

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Complex solutions

$e^{2 π i} = 1$ $e^(2pii) = 1$

So:

$2^{x} = 2^{- 5} \cdot e^{2 k π i} = 2^{- 5} \cdot 2^{\frac{2 k π}{ln (2)} i} = 2^{- 5 + \frac{2 k π}{ln (2)} i}$ $2^x = 2^-5*e^(2kpii) = 2^(-5)*2^((2kpi)/ln(2) i) = 2^(-5+(2kpi)/ln(2)i$

So $x = - 5 + \frac{2 k π}{ln (2)} i$ $x = -5 +(2kpi)/ln(2) i$ for any integer $k$ $k$

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How do you solve $2^{x} = \frac{1}{32}$ $2^x = 1/32$ ?

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