# Thread: real solutions for x^3 ln(x^2-1)-16xln(x^2-1)=0

1. ## real solutions for x^3 ln(x^2-1)-16xln(x^2-1)=0

the x cubed part is throwing me off

2. Factor out $\displaystyle \ln(x^2-1)$, then:

$\displaystyle \ln\left(x^2-1\right)\left(x^3-16x\right) = 0.$

Can you take it from there?

3. no I 'am lost

4. wait would it be

(x-1)(x+1)

(x^3-16x)
x(x^2-16x)
x(x-4)(x+4)

5. Originally Posted by Jaycs
no I 'am lost
Well if $\displaystyle \ln\left(x^2-1\right)\left(x^3-16x\right) = 0$ then either $\displaystyle \ln\left(x^2-1\right) = 0$ or $\displaystyle x^3-16x = 0$.

Originally Posted by Jaycs
wait would it be

(x-1)(x+1)

(x^3-16x)
x(x^2-16x) <=== At this step, there shouldn't be an x with the 16.
x(x-4)(x+4)
You factored both $\displaystyle x^2-1$ and $\displaystyle x^3-16x$ right! So what are the solutions for $\displaystyle x^3-16x = 0$?
For the other one, remember it's under the $\displaystyle \ln$ operator, so factoring won't do anything for now.
You need to solve $\displaystyle \ln(x^2-1) = 0,$ so remember that $\displaystyle \ln(1) = 0$ and so set $\displaystyle x^2-1 = 1$.

6. (x-4)=0
x=4

(x+4)=0
x=-4

7. Originally Posted by Jaycs
x(x-4)(x+4) = 0

(x-4)=0
x=4

(x+4)=0
x=-4
Are you done? You're missing one solution.

8. (x-1)=1
x=2
(x-1)=1
x=0

9. Originally Posted by Jaycs
(x-1)=1
x=2
(x-1)=1 , <==== + instead of -, right?
x=0
Sorry, I'm not following here. Could you explain what you're solving? If it's solving $\displaystyle x^2-1 = 1$, then
$\displaystyle (x-1)(x+1) = 1$ does not imply that either $\displaystyle (x-1) = 1$ or $\displaystyle (x+1) = 1$. Also, see the note.
You were solving x(x-4)(x+4) = 0 and you got x = 4 or x = -4 as solutions. What's the other solution?

10. is it x=0
not sure how to get the other solution if thats not it.

11. The domain demands that $\displaystyle |x| > 1$ so 0 is not a solution

From what I can see there are only two solutions: $\displaystyle x = \pm 4$

12. Originally Posted by Jaycs
is it x=0
Yes, that's it! Why the doubt? Now solve $\displaystyle x^2-1 = 1$.

13. Originally Posted by Jaycs
is it x=0
not sure how to get the other solution if thats not it.
No that gives you an undefined term: $\displaystyle \ln(-1)$.
Does $\displaystyle x=\sqrt{2}$ work?
There is one more. What?

14. Originally Posted by e^(i*pi)
The domain demands that $\displaystyle |x| > 1$ so 0 is not a solution
You're right, of course. I'm not sure whether the OP saw that x = 0 is not in the domain (I didn't for sure).
I was thinking of finding all the solutions for the x²-1 = 1 and x³-16x = 0 then discarding the unfit ones.

15. x^3-16=0
x^3=16
x=2^3

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