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

1 Answer
Jan 19, 2018

ln(e^x-1)-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(x/y) is ln(x) - ln(y)

Which simplifies to ln(e^x-1) - ln(e^x) = ln(e^x-1)-x.