Question

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

Answer 1

`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`.