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Math Help - solving exponent

  1. #1
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    solving exponent

    3\sqrt5=5^{2x}

    I thought you take the sqrt and you get 5^{1/3}=5^{2x}

    Then you get 5^3=5^{2x} So now the bases are the same and you can now solve 3=2x. Divide 2 and you get x=3/2

    The answer is not that. Where did I go wrong?
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  2. #2
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    Quote Originally Posted by Darkhrse99 View Post
    3\sqrt5=5^{2x}

    I thought you take the sqrt and you get 5^{1/3}=5^{2x}

    Then you get 5^3=5^{2x} So now the bases are the same and you can now solve 3=2x. Divide 2 and you get x=3/2

    The answer is not that. Where did I go wrong?
    Hello,

    Is it \sqrt[3]{5} or 3 \cdot \sqrt{5} ?


    There is another problem : 5^{1/3} \neq 5^3

    You have to work the same way, but with keeping 1/3.

    1/3=2x

    --> x=1/6
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  3. #3
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    Quote Originally Posted by Darkhrse99 View Post
    3\sqrt5=5^{2x}

    I thought you take the sqrt and you get 5^{1/3}=5^{2x}

    Then you get 5^3=5^{2x} So now the bases are the same and you can now solve 3=2x. Divide 2 and you get x=3/2

    The answer is not that. Where did I go wrong?
    You're going to have to use logarithms to solve this one. But first note that \sqrt{5} = 5^{0.5}. So you can divide this out and then you get 3 = 5^{(2x - 0.5)}. Take logarithms of both sides:
    \ln 3 = \ln (5^{(2x - 0.5)})

    \ln 3 = (2x - 0.5) \ln 5

    \frac{\ln 3}{\ln 5} = 2x - 0.5

    \frac{\ln 3}{\ln 5} + 0.5 = 2x

    \frac{\ln 3}{2 \ln 5} + 0.25 = x
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    Is it \sqrt[3]{5} or 3 \cdot \sqrt{5} ?


    There is another problem : 5^{1/3} \neq 5^3

    You have to work the same way, but with keeping 1/3.

    1/3=2x

    --> x=1/6
    It's \sqrt[3]{5}.
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