# How do you solve Ln(1-e^-x)?

Jan 19, 2018

$\ln \left({e}^{x} - 1\right) - x$

#### Explanation:

We can write the argument as a fraction. We first need to get rid of the negative exponent.

ln(1−e^−x) = ln(1-1/e^x) = ln((e^x-1)/e^x)

From here, you use the identity that $\ln \left(\frac{x}{y}\right) i s \ln \left(x\right) - \ln \left(y\right)$

Which simplifies to $\ln \left({e}^{x} - 1\right) - \ln \left({e}^{x}\right) = \ln \left({e}^{x} - 1\right) - x$.