How to solve for X in an ln equation

[ ln (x) ]^2- 2ln (x)

Thank you!

[ ln (x) ]^2- 2ln (x)
ln^2(x) - 2ln(x) = 0
ln(x)[ln(x) - 2] = 0
ln(x) = 0
x = e^0 = 1
ln(x) - 2 = 0
ln(x) = 2
x = e^2

[ln (x) ]^2- 2ln (x) = 0

take out ln(x) as common factor u get :

ln(x) [ln(x) - 2] = 0

either ln(x) = 0 => x = e^0 = 1 <<<<<<<<<<<

or ln(x) - 2 = 0

=> ln(x) = 2

hence x = e^2 <<<<<<<<<<