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Math Help - rearrrangement of an equation

  1. #1
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    rearrrangement of an equation

    I need to find the following equation in terms of y=f(x):

    ln(3-y)=3ln(2x+1)-x

    the bit that is stumpin me it the how to get the y out of the initial set of brackets to start the equation.

    Thanks for any help
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by wannberocketscientist View Post
    I need to find the following equation in terms of y=f(x):

    ln(3-y)=3ln(2x+1)-x

    the bit that is stumpin me it the how to get the y out of the initial set of brackets to start the equation.

    Thanks for any help
    Not a bad question:

    ln(3-y)=3ln(2x+1)-x

    Exponentiate each side:

    e^{ln(3-y)}=e^{3ln(2x+1)-x}

    3-y=e^{3ln(2x+1)}e^{-x}

    3-y=e^{ln(2x+1)^3}e^{-x}

    3-y=(2x+1)^3e^{-x}

    y=3-(2x+1)^3e^{-x}
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by wannberocketscientist View Post
    I need to find the following equation in terms of y=f(x):

    ln(3-y)=3ln(2x+1)-x

    the bit that is stumpin me it the how to get the y out of the initial set of brackets to start the equation.

    Thanks for any help
    ln(3 - y) = 3 \cdot ln(2x + 1) - x

    First, a bit of "cleaning up."

    ln(3 - y) = ln((2x + 1)^3) - x

    Now raise both sides to the base e and recall that e^{ln(z)} = z:
    3 - y = e^{ln((2x + 1)^3) - x} = (2x + 1)^3 \cdot e^{-x}

    I'm sure you can finish from here.

    -Dan
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  4. #4
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    thanks for the quick reply!
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by wannberocketscientist View Post
    thanks for the quick reply!
    You are welcome wannberocketscientist!
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