If y=1−x+x^2/(2!)−x^3/(3!)+… and z=−y−y^2/2−y^3/3−… then ln(1/(1−e^x))

Meossi91

Meossi91

Answered question

2022-08-11

If y = 1 x + x 2 2 ! x 3 3 ! + and z = y y 2 2 y 3 3 then ln ( 1 1 e x )
I can see that y = e x and z = ln ( 1 y ) . And, so z = ln ( 1 e x ). But how to go further?

Answer & Explanation

Ben Logan

Ben Logan

Beginner2022-08-12Added 13 answers

Once you know that z = log ( 1 e x ), you immediately have that
log ( 1 1 e z ) = log ( 1 1 e log ( 1 e x ) ) = log ( 1 1 ( 1 e x ) ) = log ( 1 e x ) = log e x = x
And you've done.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?