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

  1. #1
    Junior Member Jaycs's Avatar
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    real solutions for x^3 ln(x^2-1)-16xln(x^2-1)=0

    the x cubed part is throwing me off
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  2. #2
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    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?
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  3. #3
    Junior Member Jaycs's Avatar
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    no I 'am lost
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  4. #4
    Junior Member Jaycs's Avatar
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    wait would it be

    (x-1)(x+1)

    (x^3-16x)
    x(x^2-16x)
    x(x-4)(x+4)
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  5. #5
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    Quote Originally Posted by Jaycs View Post
    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.

    Quote Originally Posted by Jaycs View Post
    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.
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  6. #6
    Junior Member Jaycs's Avatar
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    (x-4)=0
    x=4

    (x+4)=0
    x=-4
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  7. #7
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    Quote Originally Posted by Jaycs View Post
    x(x-4)(x+4) = 0

    (x-4)=0
    x=4

    (x+4)=0
    x=-4
    Are you done? You're missing one solution.
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  8. #8
    Junior Member Jaycs's Avatar
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    (x-1)=1
    x=2
    (x-1)=1
    x=0
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  9. #9
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    Quote Originally Posted by Jaycs View Post
    (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?
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  10. #10
    Junior Member Jaycs's Avatar
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    is it x=0
    not sure how to get the other solution if thats not it.
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  11. #11
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    e^(i*pi)'s Avatar
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    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
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  12. #12
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    Quote Originally Posted by Jaycs View Post
    is it x=0
    Yes, that's it! Why the doubt? Now solve x^2-1 = 1.
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  13. #13
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    Quote Originally Posted by Jaycs View Post
    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?
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  14. #14
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    Quote Originally Posted by e^(i*pi) View Post
    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 x-1 = 1 and x-16x = 0 then discarding the unfit ones.
    Last edited by TheCoffeeMachine; March 6th 2011 at 09:00 AM.
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  15. #15
    Junior Member Jaycs's Avatar
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    x^3-16=0
    x^3=16
    x=2^3
    Last edited by Jaycs; March 6th 2011 at 08:58 AM. Reason: my mistake
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