How do you solve for ${log}_{x} (\frac{1}{64}) = - 3$ $log_x (1/64) = -3$ ?

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Answer 1 · 2015-11-21T05:50:33

$x = 4$ $x=4$

${log}_{x} (\frac{1}{64}) = - 3$ $log_x(1/64)=-3$

$x^{{log}_{x} (\frac{1}{64})} = x^{- 3}$ $x^(log_x(1/64))=x^-3$

$\frac{1}{64} = x^{- 3}$ $1/64=x^-3$

${(\frac{1}{64})}^{- \frac{1}{3}} = {(x^{- 3})}^{- \frac{1}{3}}$ $(1/64)^(-1/3)=(x^-3)^(-1/3)$

${(64)}^{\frac{1}{3}} = x$ $(64)^(1/3)=x$

$4 = x$ $4=x$

How do you solve for logx(164)=−3log_x (1/64) = -3?