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Math Help - Would I need to use logarithms here?

  1. #1
    Newbie
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    Smile Would I need to use logarithms here?

    If <br />
2^x4^y = 32<br />

    and <br />
\frac{3^x}{9^y}=3<br />

    then <br />
\frac{5^x}{125^y}= ?<br />


    This time the answer is 1

    I was trying to use the Logarithmic Properties to solve it, but I wasn't getting the correct answer.

    How should I solve this?

    Thanks in advance.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Medgidia, Romania
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    Is not necesarly to use logarithms.
    2^x4^y=32\Leftrightarrow 2^x2^{2y}=2^5\Leftrightarrow 2^{x+2y}=2^5\Leftrightarrow x+2y=5 (1)
    \displaystyle \frac{3^x}{9^y}=3\Leftrightarrow \frac{3^x}{3^{2y}}=3\Leftrightarrow 3^{x-2y}=3^1\Leftrightarrow x-2y=1 (2)
    Solving the system formed by (1) and (2) you get x=3, \ y=1.
    So \displaystyle \frac{5^x}{125^y}=\frac{5^3}{125}=\frac{125}{125}=  1
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  3. #3
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    Thanks a lot.
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