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(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.
