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Math Help - A little bit lost with Log solving

  1. #1
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    A little bit lost with Log solving

    Hi All,

    Bit lost here, tried all sorts of converting but can't quite work it out, just need a push in the right direction.

    (lnx)^2-ln(x^3)=10

    the square outside the brackets is confusing me.\

    Cheers,

    Perjac
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  2. #2
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    Re: A little bit lost with Log solving

    Quote Originally Posted by perjac View Post
    Hi All,

    Bit lost here, tried all sorts of converting but can't quite work it out, just need a push in the right direction.

    (lnx)^2-ln(x^3)=10

    the square outside the brackets is confusing me.\

    Cheers,

    Perjac
    Use

    ln\left(a^b\right)=blna

    for the middle term
    and have a look at what kind of resultant equation you have.
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  3. #3
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    Re: A little bit lost with Log solving

    Quote Originally Posted by perjac View Post
    (\ln(x))^2-\ln(x^3)=10
    You should know that \ln(x^3)=3\ln(x).
    Let z=\ln(x) and you get z^2-3z-10=0 to solve.

    Can you go forward?
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  4. #4
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    Re: A little bit lost with Log solving

    I guess this then becomes an algebra problem (let lnx = a)

    a^2 - 3a = 10

    though I am not sure how to solve for a. Also is there a link to how to use teh math tags?
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  5. #5
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    Re: A little bit lost with Log solving

    Solve for a using the quadratic formula.
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  6. #6
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    Re: A little bit lost with Log solving

    Duh. I shoudl have seen that.

    Cheers,

    Perjac
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  7. #7
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    Re: A little bit lost with Log solving

    Quote Originally Posted by perjac View Post
    I guess this then becomes an algebra problem (let lnx = a) a^2 - 3a = 10
    though I am not sure how to solve for a. Also is there a link to how to use teh math tags?
    I do not believe you there.
    If you cannot solve that simple quadratic equation, you have no business trying to solve this question.

    As for the use of LaTeX:
    Why not learn to post in symbols? You can use LaTeX tags
    Example
    [tex] a^2-3a=10[/tex] gives  a^2-3a=10
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  8. #8
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    Re: A little bit lost with Log solving

    Quote Originally Posted by perjac View Post
    I guess this then becomes an algebra problem (let lnx = a)

    a^2 - 3a = 10

    though I am not sure how to solve for a. Also is there a link to how to use teh math tags?
    You can just find the factors of 10 that differ by 3,
    which are 5 and 2.

    a^2-3a-10=(a-5)(a+2)

    and if this is zero, then either factor may be zero.

    Or

    a^2-3a=a(a-3)=10=5(2) \;\;or\;\; (-2)(-5)
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  9. #9
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    Re: A little bit lost with Log solving

    Quote Originally Posted by Plato View Post
    I do not believe you there.
    If you cannot solve that simple quadratic equation, you have no business trying to solve this question.

    As for the use of LaTeX:
    Why not learn to post in symbols? You can use LaTeX tags
    Example
    [tex] a^2-3a=10[/tex] gives  a^2-3a=10
    yeah brain fart there. My bad. Typed before thinking.

    Thanks for the help guys.
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  10. #10
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    Re: A little bit lost with Log solving

    Quote Originally Posted by Archie Meade View Post
    You can just find the factors of 10 that differ by 3,
    which are 5 and 2.

    a^2-3a-10=(a-5)(a+2)

    and if this is zero, then either factor may be zero.

    Or

    a^2-3a=a(a-3)=10=5(2) \;\;or\;\; (-2)(-5)

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