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Math Help - [SOLVED] How to change the base of y=A(1.3)^-t

  1. #1
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    [SOLVED] How to change the base of y=A(1.3)^-t

    The question says:
    " y=A(1.3)^{-t}
    By changing the base of the exponent to e, express y in terms of A, e and t."
    I tried:
    y=A(1.3)^{-t}\Rightarrow\frac{y}{A}=1.3^{-t} \Rightarrow log_{1.3}\frac{y}{A}=-t\Rightarrow\frac{\ln\frac{y}{A}}{\ln1.3}=-t

    \Rightarrow\ln\frac{y}{A}=-t\ln 1.3\Rightarrow\ln y-\ln A=\ln1.3^{-t}\Rightarrow\ln y=\ln A+\ln1.3^{-t}

    \Rightarrow\ln y=\ln A(1.3)^{-t}\Rightarrow y=A(1.3)^{-t}

    It just goes round in a circle. What am I doing wrong?
    Thanks in advance.
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  2. #2
    Member pflo's Avatar
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    Changing exponential bases

    Taking the last equality on your second line:
    \Rightarrow\ln\frac{y}{A}=-t\ln 1.3\Rightarrow\ln y-\ln A=\ln1.3^{-t}\Rightarrow\ln y=\ln A+\ln1.3^{-t}
    You can simplify like this:
    y=e^{ln A + ln (1.3^{-t})}=e^{lnA} * e^{-t*ln1.3}
    y=A*e^{-t*ln1.3}

    Another way of thinking of it is to solve the following for x:
    e^{x}=1.3 to get x=ln{1.3}.
    Then, instead of using 1.3 as your base, use e^{ln{1.3}}

    It may seem strange to leave your answer in this form, without simplifying the e to the "ln" power, but there could be a variety of reasons for it. For example, now you know that ln{1.3} is the rate of decay in a continuous decay model.
    Last edited by pflo; April 27th 2009 at 02:23 PM.
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  3. #3
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    Thanks a lot, that's a really great answer. : )
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