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

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I know there are 3 solutions, but I don't know the steps taken to solve it.

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Answer 1 · 2017-07-17T01:50:18

#logx^3=(logx)^3#

#=>logx^3=(logx)^3#

#=>3logx=(logx)^3#

let #logx=y#

then the equation becomes

#y^3-3y=0#

#=>y(y^2-3)=0#

When #y=0#

#=>logx=0=log1#

#=>x=1#

when

#y^2-3=0#

#=>y=pmsqrt3#

#=>logx=pmsqrt3#

#=>x=e^(pmsqrt3)#