Question 203483
log(x^2) = (logx)^2,,,,remember loga^b=b*log a
.
2logx = logx * logx
.
2 logx -logx*logx =0
.
factor
.
If 2a-a^2=0,, we  would  factor,,,a(2-a)=0
Then  we  would set,,a=0,,and  then  set (2-a) =0,,,to  solve  for  a
Well,  let  a =  "log x",,  and  repeat  the  exercise.
.
log x(2-log x) = 0
.
log x = 0,,,,,x=1
.
2-log x =0,,,2= log x,,,,x=100
.
check,,,(x=1),,,log1^2 =(log1)^2,,0=0,,,ok
.
(x=100) log100^2 = (log100)^2,,,,4=4,,,,ok