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Math Help - exponential equation help pls

  1. #1
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    exponential equation help pls

    solve the equation

    ln(2+e^-x)=2 its e to the power -x

    answer correct to 2 dp
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  2. #2
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    Hi vserian

    Hint :

    \log_a b = c

    b = a^c
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by songoku View Post
    Hi vserian

    Hint :

    \log_a b = c

    b = a^c
    And in the OP's case a = e

    Also a^{log_a(b)} = b
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  4. #4
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    Quote Originally Posted by vserian View Post
    solve the equation

    ln(2+e^-x)=2 its e to the power -x

    answer correct to 2 dp
    This equation can be solved exactly:

    \ln(2+e^{-x})=2

    \ln \left(\dfrac{2e^x+1}{e^x} \right)=2

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

    \ln (2e^x+1) =x + 2

    2e^x+1 =e^{x + 2} = e^2 \cdot e^x

    1 = e^2 \cdot e^x - 2e^x = e^x(e^2-2)

    \dfrac1{e^2-2} =  e^x

    -\ln(e^2-2) =  x
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