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Math Help - Simplifying Logs

  1. #1
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    Smile Simplifying Logs

    Hi there, I am a bit stuck as to how to do this question. I know the laws of logs but this has me stumped!

    Thanks in advance

    Express the following without logarithms:

    lnP=1/2ln(Q+1)-3lnR+2

    Thanks again for any input

    Mike
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  2. #2
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    Quote Originally Posted by mikewhant View Post
    Hi there, I am a bit stuck as to how to do this question. I know the laws of logs but this has me stumped!

    Thanks in advance

    Express the following without logarithms:

    lnP=1/2ln(Q+1)-3lnR+2

    Thanks again for any input

    Mike
    Its
    <br />
\frac{1}{2ln(Q+1)-3ln(R)+2}<br />
    or
    <br />
\frac{1}{2ln(Q+1)}-3ln(R)+2<br />

    Using brackets is nicely to make something readable.

    **Edit**:
    OR
    <br />
\frac{1}{2}ln(Q+1)-3ln(R)+2<br />
    when Amer solve your question according to it?
    Last edited by TWiX; January 9th 2010 at 10:56 AM.
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by mikewhant View Post
    Hi there, I am a bit stuck as to how to do this question. I know the laws of logs but this has me stumped!

    Thanks in advance

    Express the following without logarithms:

    lnP=1/2ln(Q+1)-3lnR+2

    Thanks again for any input

    Mike
    you know that

    \log a + \log b = \log ab

    \log a - \log b = \log \frac{a}{b}

    \ln e = 1

    a\log b = \log b^a

    \ln P = 1/2 \ln (Q+1) - 3\ln R + 2

     \ln P = \ln (Q+1)^{\frac{1}{2}} - \ln R^3 + 2 \ln e

    \ln P = \ln \frac{(Q+1)^{\frac{1}{2}}}{R^3} + \ln e^2

    \ln P = \ln \left(\frac{(Q+1)^{\frac{1}{2}}}{R^3} \cdot e^2 \right)

    P = \frac{(Q+1)^{\frac{1}{2}}}{R^3} \cdot e^2
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  4. #4
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    Smile

    Thats great thanks alot.

    Take care

    Mike
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