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Math Help - Logarithmic Troubles

  1. #1
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    Logarithmic Troubles

    The Equation is  x(y) = 1000e^{1.3y} x = 2000 find y
    So applied Logarithms to both sides which left me:


     <br />
\log_{10} 2000 + \log_{10} y = \log_{10} 1000 + 1.3y \log_{10} e<br />
     <br />
3.30103 + \log_{10} y= 3 + 1.3y(0.43429)<br />
     <br />
\log_{10} y = 3 + 0.56458y -3.30103<br />
     <br />
\log_{10} y = 0.56458y -0.30103<br />

    I'm not sure where to go from here?
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Jbuckham View Post
    The Equation is  x(y) = 1000e^{1.3y} x = 2000 find y
    So applied Logarithms to both sides which left me:

     <br />
\log_{10} 2000 + \color{red}\log_{10} y\color{black} = \log_{10} 1000 + 1.3y \log_{10} e<br />
    Where did that come from? Proceed instead as follows.

    1000e^{1.3y}=2000

    \Rightarrow\ln\left(1000e^{1.3y}\right)=\ln2000

    \Rightarrow\ln1000+\ln e^{1.3y}=\ln2000

    \Rightarrow1.3y=\ln2000-\ln1000

    \Rightarrow1.3y=\ln\frac{2000}{1000}

    \Rightarrow y=\frac{10}{13}\ln2
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  3. #3
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    Quote Originally Posted by Reckoner View Post
    Where did that come from? Proceed instead as follows.

    The original equation is  x \color{red}\ (y) \color{black}  = 1000e^{1.3y}
    therefore Log x + Log y = log 1000 + 1.3ylog e or 1000e^{1.3y}=2000y
    Thats where that log y came from...
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  4. #4
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    Hello Jbuckham
    Quote Originally Posted by Jbuckham View Post
    The original equation is  x \color{red}\ (y) \color{black}  = 1000e^{1.3y}
    therefore Log x + Log y = log 1000 + 1.3ylog e or 1000e^{1.3y}=2000y
    Thats where that log y came from...
    I think it's a simple case of not interpreting the notation correctly.

    You have interpreted x(y) to mean x \times y. But I think that, in this question, x(y) = ... is function notation, and means x, expressed as a function of y, is equal to ...

    So the equation you're being asked to solve is simply:

    x = 1000e^{1.3y}

    (Why do people make life more difficult than it needs to be, eh?)

    Grandad
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  5. #5
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    Quote Originally Posted by Jbuckham View Post
    The original equation is  x \color{red}\(y)\color{black}  = 1000e^{1.3y}
    therefore Log x + Log y = log 1000 + 1.3ylog e
    Thats where that log y came from...
    Oh, I thought when you wrote x(y) that you were specifying x as a function of y. Instead, we have

    1000e^{1.3y}=2000y,

    which has no solutions. You can see from their graphs: the exponential function starts out at much higher values than the linear function, and then it begins to grow at such a quickly increasing rate that the linear function can never hope to catch up.

    Edit: Grandad may be right--perhaps you have misinterpreted the problem? You may want to state the question exactly as it is written in your book or assignment.
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  6. #6
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    You're right I have misinterpreted the problem. Thank you very much for your help. it had me completly stumpted
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  7. #7
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    Quote Originally Posted by Jbuckham View Post
    The original equation is  x \color{red}\ (y) \color{black}  = 1000e^{1.3y}
    therefore Log x + Log y = log 1000 + 1.3ylog e...
    Those parentheses on the left-hand side of the "equals" sign do not indicate multiplication!

    To learn how to work with "function notation", try here.
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