logx^2除log1/x是=幾?

logx^2=2logx (1)
log1/x=logx^-1=-logx (2)
Therefore,
logx^2除log1/x=2logx除(-logx)
=-2

I hope you understand the logic or principle from the equation 1 and 2
That is, logx^y=ylogx

logx^2=2logx---------------------(1)
log1/x=logx^(-1)= -logx--------(2)
原式=1式除2式= -2