0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ if there is the modulus (absolute) sign, would 1/3 be the answer too? $\endgroup$ – sam May 15 at 9:03
  • $\begingroup$ For $$\ln(x^{1/3})$$ we nead no absolute signs, if we define $$x>0$$ $\endgroup$ – Dr. Sonnhard Graubner May 15 at 9:08
  • $\begingroup$ Thanks for reply. Sorry. I am just a bit confused. Can x = -1 negative number for y = ln x^2? $\endgroup$ – sam May 15 at 9:21
  • $\begingroup$ Yes it nis $$\ln((-1)^2)=\ln(1)=0$$ $\endgroup$ – Dr. Sonnhard Graubner May 15 at 9:26
  • $\begingroup$ 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. $\endgroup$ – sam May 15 at 9:34
0
$\begingroup$

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:

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.