Solve logarithm equation?

log x^2 = (log x)^2

2log x = log x logx

2 = log x logx / logx

Cancel down.

2 = log x

Antilog.

x = 10^2

x = 100

  Note:  ( log(x) )²  is normally written as:  log ² (x)

    log( x² ) = log ² (x)   ...  assuming base ten logs.

   2 log(x) = log ² (x)

           0 = log ² (x)  –  2 log(x)

           0 = log(x) • [ log(x)  –  2 ]

      log(x) = 0       and       log(x)  –  2 = 0

            x = 1           ...             log(x) = 2

                          ...                       x = 10² = 100

So there are two solutions that are valid in the initial log equation

as shown below:

    log( x² ) = log ² (x)

    log( 1² ) = log ² (1)   ...  x = 1   solution

     log( 1 ) = log ² (1)

            0 = (0)²   ...  TRUE

          log( x² ) = log ² (x)

        log( 100² ) = log ² (100)   ...  x = 100   solution

    log( 10000 ) = log ² (100)

                  4 = (2)²   ...  TRUE

First, log x^2 = 2log x so now we have 2log x = (log x)^2 so to simplify the writing I'll write L for log x so we now have 2L = l^2 so L^2 - 2L = 0 giving l(L - 2) = 0 so L = 0 or L = 2 i.e. log x = 0 or log x = 2 and that gives us x= 1 or x = 100.