# exponential equation help pls

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• Aug 16th 2009, 06:52 AM
vserian
exponential equation help pls
solve the equation

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

answer correct to 2 dp
• Aug 16th 2009, 06:54 AM
songoku
Hi vserian

Hint :

$\displaystyle \log_a b = c$

$\displaystyle b = a^c$
• Aug 16th 2009, 08:28 AM
e^(i*pi)
Quote:

Originally Posted by songoku
Hi vserian

Hint :

$\displaystyle \log_a b = c$

$\displaystyle b = a^c$

And in the OP's case a = e

Also $\displaystyle a^{log_a(b)} = b$
• Aug 16th 2009, 10:54 AM
earboth
Quote:

Originally Posted by vserian
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:

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

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

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

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

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

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

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

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