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Thread: Questions on logarithms

  1. #1
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    Questions on logarithms

    solving P and Q where they are integers

    (25/16)^ -3/2 = 2^P * 5^Q
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    Re: Questions on logarithms

    Quote Originally Posted by Britt00 View Post
    solving P and Q where they are integers
    (25/16)^ -3/2 = 2^P * 5^Q
    ${\left( {\dfrac{{25}}{{16}}} \right)^{\dfrac{{ - 3}}{2}}} = {\left( {\dfrac{{{5^2}}}{{{2^4}}}} \right)^{\dfrac{{ - 3}}{2}}} = \dfrac{{{5^{ - 3}}}}{{{2^{ - 6}}}} = ?$
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    Re: Questions on logarithms

    no need for logs, note ...

    $\left(\dfrac{25}{16}\right)^{-3/2} = \left(\dfrac{16}{25}\right)^{3/2} = \left(\dfrac{4^2}{5^2}\right)^{3/2} = \left(\dfrac{4}{5}\right)^3 = 4^3 \cdot 5^{-3}$

    can you finish finding the values of P and Q from here?
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    Re: Questions on logarithms

    Is it p=6 and q=-3 ?
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    Re: Questions on logarithms

    Quote Originally Posted by Britt00 View Post
    Is it p=6 and q=-3 ?
    Yes.
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    Re: Questions on logarithms

    Quote Originally Posted by Britt00 View Post
    Is it p=6 and q=-3 ?
    Substitute back into the original equation to check on your own.

    $\left ( \dfrac{25}{16} \right )^{-(3/2)} = \left ( \dfrac{16}{25} \right )^{(3/2)} = \left ( \sqrt{\dfrac{16}{25} } \right )^3 =$

    $\left ( \dfrac{4}{5} \right )^3 = \left ( \dfrac{2^2}{5} \right )^3 = \dfrac{2^6}{5^3} = 2^6 * 5^{-3}.$
    Last edited by JeffM; Mar 5th 2017 at 06:58 PM.
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