How do you evaluate ${log}_{x} (\frac{1}{36}) = - (\frac{2}{3})$ $log_x(1/36)= -(2/3)$ ?

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How do you evaluate ${log}_{x} (\frac{1}{36}) = - (\frac{2}{3})$ $log_x(1/36)= -(2/3)$ ?

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Answer 1 · 2016-03-10T21:53:33

$x = 216$ $x=216$

Explanation:

$x^{- \frac{2}{3}} = \frac{1}{36}$ $x^(-2/3) =1/36$
$x^{- \frac{2}{3}} = 6^{- 2}$ $x^(-2/3) =6^-2$
${(x^{- \frac{2}{3}})}^{- \frac{3}{2}} = {(6^{- 2})}^{- \frac{3}{2}}$ $(x^(-2/3))^(-3/2) =(6^-2)^(-3/2)$
$x = 6^{3} = 216$ $x=6^3=216$

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How do you evaluate ${log}_{x} (\frac{1}{36}) = - (\frac{2}{3})$ $log_x(1/36)= -(2/3)$ ?

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Explanation:

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