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Math Help - graphing problem

  1. #1
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    graphing problem

    i have this equation:

    solve for x:
    ln(x)+ln(x-3)=2

    x=0, x=3

    In addition:

    graph f(x)=ln(x) and -f(x-3)+2 by using transformations. Explain the relationship between the solutions to the original equation and the points of intersection on your graph.

    How do you graph -f(x-3)+2?

    please help Idon't understand the problem. thanks!
    Last edited by Anemori; May 19th 2010 at 03:55 AM.
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  2. #2
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    Quote Originally Posted by Anemori View Post
    i have this equation:

    solve for x:
    ln(x)+ln(x-3)=2

    x>0 and x>3 <<<<<<<<< typo
    Use the laws of logarithms:

    \ln(x)+\ln(x-3)=2~\implies~\ln(x(x-3))=\ln(e^2)

    Now use the logarithms as exponents to the base e. You'll get a quadratic equation. Keep in mind that this equation is only valid for x > 3


    In addition:

    graph f(x)=ln(x) and -f(x-3)+2 by using transformations. Explain the relationship between the solutions to the original equation and the points of intersection on your graph.

    How do you graph -f(x-3)+2?

    please help Idon't understand the problem. thanks!
    1. Graph f(x)=\ln(x)

    2. f(x-3) means the graph of f is translated by 3 units to the right.

    3. -f(x-3) means the translated graph is refelcted about the x-axis.

    4. -f(x-3) + 2 means the translated and reflected graph is translated by 2 units upwards.

    5. The x-coordinate of the point of intersection is the solution to the equation above:

    \ln(x)+\ln(x-3)=2~\implies~\underbrace{\ln(x)}_{f(x)}=\underbra  ce{-\ln(x-3)+2}_{-f(x-3)+2}
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