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

**Jaycs** · March 6th 2011, 07:20 AM

the x cubed part is throwing me off

**TheCoffeeMachine** · March 6th 2011, 07:23 AM

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

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

Can you take it from there?

**Jaycs** · March 6th 2011, 07:41 AM

no I 'am lost

**Jaycs** · March 6th 2011, 07:44 AM

wait would it be

(x-1)(x+1)

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

**TheCoffeeMachine** · March 6th 2011, 07:52 AM

Originally Posted by Jaycs

no I 'am lost

Well if $\ln\left(x^2-1\right)\left(x^3-16x\right) = 0$ then either $\ln\left(x^2-1\right) = 0$ or $x^3-16x = 0$ .

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 $x^2-1$ and $x^3-16x$ right! So what are the solutions for $x^3-16x = 0$ ?
For the other one, remember it's under the $\ln$ operator, so factoring won't do anything for now.
You need to solve $\ln(x^2-1) = 0,$ so remember that $\ln(1) = 0$ and so set $x^2-1 = 1$ .

**Jaycs** · March 6th 2011, 08:02 AM

(x-4)=0
x=4

(x+4)=0
x=-4

**TheCoffeeMachine** · March 6th 2011, 08:05 AM

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.

**Jaycs** · March 6th 2011, 08:08 AM

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

**TheCoffeeMachine** · March 6th 2011, 08:14 AM

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 $x^2-1 = 1$ , then
$(x-1)(x+1) = 1$ does not imply that either $(x-1) = 1$ or $(x+1) = 1$ . 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?

**Jaycs** · March 6th 2011, 08:37 AM

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

**e^(i*pi)** · March 6th 2011, 08:41 AM

The domain demands that $|x| > 1$ so 0 is not a solution

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

**TheCoffeeMachine** · March 6th 2011, 08:42 AM

Originally Posted by Jaycs

is it x=0

Yes, that's it! Why the doubt? Now solve $x^2-1 = 1$ .

**Plato** · March 6th 2011, 08:45 AM

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: $\ln(-1)$ .
Does $x=\sqrt{2}$ work?
There is one more. What?

**TheCoffeeMachine** · March 6th 2011, 08:45 AM

Originally Posted by e^(i*pi)

The domain demands that $|x| > 1$ 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.

**Jaycs** · March 6th 2011, 08:54 AM

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

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