Calculus: Simplifying Logs Help?

Question

Let f(x) = 1/(1-e^x) and g(x) = ln(x-1)

Find g(f(x)) and simplify your result.

Can anyone help with me with this please?

Answer 1

g(f(x)) means to substitute in f(x) for x in g(x), so

g(f(x)) = ln(1/(1-e^x) - 1). Find the common denominator, so that you have a single numerator and denominator (in this case, -1 = -1(1-e^x)/(1-e^x) which yields

ln[(1 - 1 - e^x)/(1 - e^x)] which of course simplifies to ln[e^x/(1 - e^x)]. Use the logarithmic identity ln(x/y) = ln x - ln y:

ln(e^x) - ln(1 - e^x) = x - ln(1 - e^x).

Answer 2

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

ln((1-1+e^x)/(1-e^x))

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

ln(e^x)-ln(1-e^x)

x-ln(1-e^x)

not sure if you can do anything from there.

make it a good day

Answer 3

g(f(x))= ln ((1/(1-e^x))-1)=ln ((1-(1-(e^x)))/(1-e^x))=

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