How do you solve for $log_x (1/64) = -3$ ?

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Answer 1 · 2015-11-21T05:50:33

$x=4$

$log_x(1/64)=-3$

$x^(log_x(1/64))=x^-3$

$1/64=x^-3$

$(1/64)^(-1/3)=(x^-3)^(-1/3)$

$(64)^(1/3)=x$

$4=x$

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