# find the x-intercept of a natural logarithm function with a power of 2

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?

It is $$y=2\ln|3x-2|=0$$ so $$\ln|3x-2|=\ln(1)$$ so you have to solve $$|3x-2|=1$$

• if there is the modulus (absolute) sign, would 1/3 be the answer too? – sam May 15 at 9:03
• For $$\ln(x^{1/3})$$ we nead no absolute signs, if we define $$x>0$$ – Dr. Sonnhard Graubner May 15 at 9:08
• Thanks for reply. Sorry. I am just a bit confused. Can x = -1 negative number for y = ln x^2? – sam May 15 at 9:21
• Yes it nis $$\ln((-1)^2)=\ln(1)=0$$ – Dr. Sonnhard Graubner May 15 at 9:26
• is it correct to say if x stand by itself like ln x^2, you can find the square of the x first. For a function like 3x -2 in ln (3x -2)^2, you have to move the 2 in front of ln first. – sam May 15 at 9:34

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:

$$e^{ln(3x-2)}=e^0$$

$$3x-2=1 \rightarrow x=1$$