Dumb question

1. Jan 27, 2005

physicsss

I've been trying to do this for almost an hour now but I'm still not able to show that [1-e^(-x)]/[1-e^(x)]=-e^(-x).

Thank you.

2. Jan 27, 2005

dextercioby

U mean show that
$$\frac{1-e^{-x}}{1-e^{x}}=-e^{-x}$$

Daniel.

3. Jan 27, 2005

physicsss

No, no. I got [1-e^(-x)]/[1-e^(x)] as an answer to a problem, but in the back of the book the answer is given as -e^(-x). What, I'm asking then, is how to get from my answe to the book's...I know for a fact that they're equal.

Sorry for the confusion.

4. Jan 27, 2005

dextercioby

How about multiplying both the numerator and the denominator by $e^{x}$ ??

Daniel.

5. Jan 27, 2005

physicsss

It doesn't work.

6. Jan 27, 2005

dextercioby

It works for me
$$(\frac{1-e^{-x}}{1-e^{x}})(\frac{e^{x}}{e^{x}})=(\frac{e^{x}-1}{1-e^{x}})(\frac{1}{e^{x}})=-1\cdot \frac{1}{e^{x}}=-e^{-x}$$

Daniel.

7. Jan 27, 2005

Eratosthenes

(1 - e^(-x))/(1 - e^x) = (1 - 1/e^x)/(1 - e^x) ...then find a common denominator on the top fraction and you get [(e^x - 1)/e^x]/(1 - e^x) = (e^x - 1)/[(e^x)(1 - e^x)] then -1*(1-e^x)/[(e^x)(1 - e^x)] = -1/e^x = -e^(-x)

wow i typed that fast, should be right,i included lots of steps, i skipped a few, but included more than what you'd need to write down probably, do you see it now?

8. Jan 27, 2005

physicsss

@dextercioby:

how did you turn (e^x-1)/(1-e^x) into 1? If you multiply it by -1 on the top don't you have to do it to the bottom too?

9. Jan 27, 2005

dextercioby

I didn't...I turned into "-1"...

Daniel.

10. Jan 27, 2005

Edgardo

Hello Physicsss,

by [1-e^(x)].

You get

[1-e^(-x)] = -e^(-x) * [1-e^(x)]
=> [1-e^(-x)] = -e^(-x) + (-e^(-x))*(-e^(x))
=>[1-e^(-x)] = -e^(-x) + 1
=> [1-e^(-x)] = 1-e^(-x)