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Math Help - Two Exponential Problems

  1. #1
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    Two Exponential Problems

    I don't know how to solve these two problems:

    1.) 100 = 5000(0.02^{0.5^x})

    2.) 3^x(1/9)^{x+1} = 27(3^{-x})

    Your assistance is much appreciated.
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    #1: I am going to use fractions instead of decimals.

    \frac{1}{50}=(\frac{1}{50})^{\frac{1}{2}x}

    log of both sides:

    log(\frac{1}{50})=\frac{1}{2}xlog(\frac{1}{50})

    1=\frac{1}{2}x

    x=2

    Use your log properties.
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    For #2

    <br />
3^x(1/9)^{x+1} = 27(3^{-x})<br />

    Since 9 and 27 are both powers of 3 rewrite them as their prime factorization.

    \frac{1}{9}=3^{-2} and 27=3^3

    Then we get

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

    Using exponent rules we get

    3^x \cdot (3^{-2x-2})=(3^{-x+3})

    (3^{-x-2})=(3^{-x+3})

    since the bases are the same the exponents must be equal so

    -x+2=-x+3 this equation has no solutions. Therefore this equation has no solutions.
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