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Thread: Logarithmic

  1. #1
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    Logarithmic

    Solve .
    Could someone help explain how to solve this? Cheers
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Haris View Post
    Solve .
    Could someone help explain how to solve this? Cheers
    Multiply both sides by $\displaystyle 6^x$
    Then let $\displaystyle X=6^x$ and solve the quadratic equation
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by Haris View Post
    Solve .
    If you are supposed to know what $\displaystyle \sinh$ is you can claim that

    $\displaystyle 6^x-6^{-x}=\frac{15}{4} \Longleftrightarrow \mathrm{e}^{x\ln 6}-\mathrm{e}^{-x\ln 6}=\frac{15}{4} \Longleftrightarrow \sinh(x\ln6)=\frac{15}{8}$.

    And this equation can easily be solved using the inverse function of $\displaystyle \sinh$.
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  4. #4
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    I got x=1+sqrt(20)/2 and x=1-sqrt(20)/2

    Is that correct?
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  5. #5
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    Follow Moo's advice.
    $\displaystyle \begin{gathered}
    4\left( {6^{2x} } \right) - 15\left( {6^x } \right) - 4 = 0 \hfill \\
    \left[ {4\left( {6^x } \right) + 1} \right]\left[ {\left( {6^x } \right) - 4} \right] = 0 \hfill \\ \left( {6^x } \right) \ne \frac{{ - 1}}
    {4}\,\& \,\left( {6^x } \right) = 4 \hfill \\
    \end{gathered} $
    Can you finish?
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  6. #6
    Junior Member ursa's Avatar
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    I got x=1+sqrt(20)/2 and x=1-sqrt(20)/2

    Is that correct?

    this is not the right answer
    check again
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  7. #7
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    x=ln(4)/ln(6) ?
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  8. #8
    Junior Member ursa's Avatar
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    x=ln(4)/ln(6) ?
    yup...that's it
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