find the $x$-intercept of the function $y = \ln((3x-2)^2)$.
in order to find it, move the power 2 in front of the natural log: $y = 2 \ln(3x-2)$.
for x -intercept, $y = 0$. Therefore, $\ln(3x-2) = 0$. Hence, $(3x-2) = 1$. and x = 1.
The question is why cannot set $(3x-2)^2 = 1$. This give an additional answer ($x = 1/3$), which is wrong. Why?