How do you solve $log(x^3)=(logx)^3$ ?

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Answer 1 · 2015-09-11T06:35:43

x=1, $10^sqrt3, 10^-sqrt3$

$log(x^3)= 3log x$

$3log x= (logx)^3$
$(log x)^3 -3logx =0$

$logx ( (log x)^2-3)$ =0

logx=0, $(logx)^2 -3=0$

log x=0 means x=1 and $(logx )^2 -3=0$ would mean $log x=+-sqrt3$ , or $x=10^sqrt3, 10^-sqrt3$

How do you solve log(x^3)=(logx)^3log(x^3)=(logx)^3?