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Math Help - Solve equation involving logarithms

  1. #1
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    Solve equation involving logarithms

    Hi All,

    I was studying logarithms today, and came across this problem. Need a hint on how to proceed.

    2^{4x} - 3^{3x} + 4^{2x} = 0

    I simplified it down to,

    \frac{2^{4x}}{3^{3x}} = \frac{1}{2}

    How do I proceed further? Thanks!
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  2. #2
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    Quote Originally Posted by mathguy80 View Post
    Hi All,

    I was studying logarithms today, and came across this problem. Need a hint on how to proceed.

    2^{4x} - 3^{3x} + 4^{2x} = 0

    I simplified it down to,

    \frac{2^{4x}}{3^{3x}} = \frac{1}{2}

    How do I proceed further? Thanks!
    I take your last line:

    \dfrac{2^{4x}}{3^{3x}} = \dfrac{16^x}{27^x} = \left(\dfrac{16}{27}  \right)^x = \dfrac{1}{2}

    Now use logarithms to the base \frac{16}{27}
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  3. #3
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    Thanks @earboth. Looks simple now!
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