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

1 Answer
Feb 25, 2017

We can write the argument as a fraction after getting rid of the negative exponent:

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

From here, use #ln(a/b)=ln(a)-ln(b)# and #ln(e^x)=x#:

#=ln(e^x-1)-ln(e^x)=color(blue)(ln(e^x-1)-x#

I don't know if this is a simplification per se, but it's definitely a valid way to rewrite the function.