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Math Help - Logs

  1. #1
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    Logs

    So I have this equation:

    (1/27)^(x-1) = 3 times cubed root of nine

    I need to solve for x this using logs...

    Any thoughts?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by RFB4191 View Post
    So I have this equation:

    (1/27)^(x-1) = 3 times cubed root of nine

    I need to solve this using logs...

    Any thoughts?
    \bigg(\frac{1}{27}\bigg)^{x-1}=\frac{1}{27^{x-1}}=\frac{1}{3^{3x-3}}=3^{-(3x-3)}

    also note that 3\cdot\sqrt[3]{9}=3\cdot\sqrt[3]{3^2}=3\cdot{3^{\frac{2}{3}}}=3^{\frac{5}{3}}

    So now we have 3^{-(3x-3)}=3^{\frac{5}{3}}

    Taking the \log_3 of both sides we get

    -(3x-3)=\frac{5}{3}

    I think you can take it from there
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by RFB4191 View Post
    So I have this equation:

    (1/27)^(x-1) = 3 times cubed root of nine

    I need to solve for x this using logs...

    Any thoughts?
    \left ( \frac{1}{27} \right ) ^{x - 1} = 3 \sqrt[3]{9}

    27^{-(x - 1)} = 3 \cdot 3^{2/3}

    3^{-3(x - 1)} = 3^{5/3}

    Can you take it from here?

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    \left ( \frac{1}{27} \right ) ^{x - 1} = 3 \sqrt[3]{9}

    27^{-(x - 1)} = 3 \cdot 3^{2/3}

    3^{-3(x - 1)} = 3^{5/3}

    Can you take it from here?

    -Dan

    I got it... thanks a lot
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