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Math Help - logarithms(solve for x)

  1. #1
    VkL
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    logarithms(solve for x)

    4 * 2^x = 9 * 6^x

    I took ln of both sides, and couldent get anywhere=/ Can any1 help?
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  2. #2
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    Quote Originally Posted by VkL View Post
    4 * 2^x = 9 * 6^x

    I took ln of both sides, and couldent get anywhere=/ Can any1 help?
    4 \cdot 2^x = 9 \cdot 6^x

    \frac{4}{9} = \frac{6^x}{2^x}

    \frac{4}{9} = \left(\frac{6}{2}\right)^x

    \frac{4}{9} = 3^x

    4 = 9 \cdot 3^x

    4 = 3^{x+2}

    \ln(4) = (x+2)\ln(3)

    \frac{\ln{4}}{\ln{3}} = x + 2

    x = \frac{\ln{4}}{\ln{3}} - 2
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  3. #3
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    Quote Originally Posted by VkL View Post
    4 * 2^x = 9 * 6^x
    Simple way:

    6^x / 2^x = 4/9

    (6/2)^x = 4/9

    3^x = 4/9

    x = log(4/9) / log(3)
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