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Math Help - Logarithms

  1. #1
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    Logarithms

    Hi,

    \text{If} \log 4 + 2 \log p = 2, \,\,\text{find p.}

    Initially I tried this:

    \log\, 4 + \log p^2 = 2

    \log\, 4p^2 = \log 2\bigg]\Rightarrow logs cancel since they're of the same base

    4p^2 = 2

    p = \sqrt{\frac{1}{2}}

    I found my answer to be incorrect. I want to know why.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Hellbent View Post
    Hi,

    \text{If} \log 4 + 2 \log p = 2, \,\,\text{find p.}

    Initially I tried this:

    \log\, 4 + \log p^2 = 2

    \log\, 4p^2 = \log 2\bigg]\Rightarrow logs cancel since they're of the same base
    Why did you introduce a log on the right hand side? There shouldn't be one there.


    Your third line should read \log(4p^2) = 2

    So you get 4p^2 = 10^2 in base 10 or 4p^2 = e^2 in base e


    edit: you touched upon it but remember to discard the negative value of p since p>0

    edit 2: I get an answer of p=5
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  3. #3
    MHF Contributor Unknown008's Avatar
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    This line is not true.

    \log\, 4p^2 = \log 2

    This is true:

    \log\, 4p^2 = 2
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  4. #4
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    Spoiler:
    \log{4} + 2\log p = 2 \Rightarrow \log{2^2} + 2\log{p} = 2 \Rightarrow 2\log{2} + 2\log{p} = 2 \Rightarrow  2\left(\log{2}+\log{p}\right) = 2.
    Spoiler:
    So \log{2}+\log{p} = 1, therefore \log{p} = 1-\log{2}, just put p on one side.
    Spoiler:
    \displaystyle \log{p} = 1-\log{2} \Leftrightarrow e^{\log{p}} = e^{1-\log{2}} \Rightarrow p = e \left(e^{-\log(2)\right) = \frac{e}{e^{\log{2}}} = \frac{e}{2}.

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