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Thread: How would I simplify this?

  1. #1
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    How would I simplify this?

    $\displaystyle Log\sqrt{x^3-3x}=\frac{1}{2}$

    I got to: $\displaystyle 10^\frac{1}{2}=(x^3-3x)^\frac{1}{2}$ and then got a little confused on how I would distribute the exponent and solve it.

    Any and all pointers highly appreciated, thanks!
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  2. #2
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    e^(i*pi)'s Avatar
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    Using Log laws:

    $\displaystyle \log(\sqrt{x^3-3x}) = \dfrac{1}{2}\log(x^3-3x)$

    $\displaystyle \log(x^3-3x) = 1$

    $\displaystyle x^3-3x = 10$

    Can you continue? Bear in mind $\displaystyle x > 0$

    ===================

    Using your method you can square both sides to give $\displaystyle 10 = x^3-3x$ which goes to show there's more than one way to skin a cat. Since we know that $\displaystyle x>0$ we needn't worry about extraneous solutions
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Using Log laws:

    $\displaystyle \log(\sqrt{x^3-3x}) = \dfrac{1}{2}\log(x^3-3x)$

    $\displaystyle \log(x^3-3x) = 1$
    Hmm, do you mind showing me how you got to $\displaystyle \log(x^3-3x) = 1$?

    Quote Originally Posted by e^(i*pi) View Post
    Using your method you can square both sides to give $\displaystyle 10 = x^3-3x$ which goes to show there's more than one way to skin a cat. Since we know that $\displaystyle x>0$ we needn't worry about extraneous solutions
    Ah, that would work. Thanks a lot!
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by youngb11 View Post
    Hmm, do you mind showing me how you got to $\displaystyle \log(x^3-3x) = 1$?



    Ah, that would work. Thanks a lot!
    Certainly

    You have the equation $\displaystyle \log(\sqrt{x^3-3x}) = \dfrac{1}{2}$. By the laws of logarithms $\displaystyle [\ln(a^k) = k\ln(a)]$ so it can be as $\displaystyle \dfrac{1}{2}\log(x^3-3x) = \dfrac{1}{2}$.

    Either by inspection or multiplying by 2 will cancel the 1/2 on both sides to give $\displaystyle \log(x^3-3x) = 1$
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    Certainly

    You have the equation $\displaystyle \log(\sqrt{x^3-3x}) = \dfrac{1}{2}$. By the laws of logarithms $\displaystyle [\ln(a^k) = k\ln(a)]$ so it can be as $\displaystyle \dfrac{1}{2}\log(x^3-3x) = \dfrac{1}{2}$.

    Either by inspection or multiplying by 2 will cancel the 1/2 on both sides to give $\displaystyle \log(x^3-3x) = 1$
    Thanks a lot for the clear explanation! Have a great day
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