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Math Help - logarithmics

  1. #1
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    logarithmics

    finding all solutions to this question.


    <br />
2\log y = log(3-2y)<br />

    <br />
2\log y = log(\frac {3} {2y})<br />

    <br />
2\log y = \frac {3} {2y}<br />

    Now.. im not to sure what to do with the
    <br />
2\log y<br />

    thank u!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jvignacio View Post
    finding all solutions to this question.


    <br />
2\log y = log(3-2y)<br />

    <br />
2\log y = log(\frac {3} {2y})<br />

    <br />
2\log y = \frac {3} {2y}<br />

    Now.. im not to sure what to do with the
    <br />
2\log y<br />

    thank u!
    Woah! We have got some problems here

    2\log_{10}(y)=\log_{10}(y^2)=\log_{10}(3-2y)\Rightarrow\log_{10}(y^2)-\log_{10}(3-2y)=\log_{10}\left(\frac{y^2}{3-2y}\right)=0

    Now exponentiating both sides we get

    \frac{y^2}{3-2y}=1\Rightarrow{y^2-3+2y=0}\Rightarrow{y=\frac{-2\pm\sqrt{4-4(-3)(1)}}{2}}=-1\pm{2}

    One of which is an extraneous solution
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  3. #3
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    just wondering why u made the last line start off with equaling to 1 when you took the log off to find its y value using quadratics and when you have the log base 10 and equation you had it equal to 0.

    thanks
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jvignacio View Post
    just wondering why u made the last line start off with equaling to 1 when you took the log off to find its y value using quadratics and when you have the log base 10 and equation you had it equal to 0.

    thanks
    if \log_{10}(u(x))=0\Rightarrow{u(x)=10^0=1}
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