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

1 Answer
Jan 19, 2018

Answer:

#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#.