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Thread: 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 :

    $\displaystyle \log_a b = c$

    $\displaystyle b = a^c$
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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 :

    $\displaystyle \log_a b = c$

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

    Also $\displaystyle a^{log_a(b)} = b$
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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:

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

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

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

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

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

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

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

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