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?
On
We have $y=\ln(3x-2)^2$, since by the power of logs we can bring the power down as a coefficient of the log we get $y=2\ln(3x-2)$. For a $x$-intercept we have $y=0$ so $2\ln(3x-2)=0$, then $\ln(3x-2)=0$.
Next we can take the exponentials of each side:
$$e^{ln(3x-2)}=e^0$$
$$3x-2=1 \rightarrow x=1$$
It is $$y=2\ln|3x-2|=0$$ so $$\ln|3x-2|=\ln(1)$$ so you have to solve $$|3x-2|=1$$