# Exponential question, x = [ 1 / 1 - e^-1.768] x e^-0.884

1. Aug 27, 2007

### nophysics

Im trying to setup this equation so I can figure it out, I have tried a couple of ways but I have lost the fundamental rules years ago - can anyone help me set it up for me ? Thanks so much.

Equation: x = [ 1 / 1 - e^-1.768]e^-0.884
; Solve for x

Last edited: Aug 28, 2007
2. Aug 27, 2007

### ZioX

Is that a typo with the second x? What is that second x doing?

If there is no typo, then x=ax so that x(1-a)=0 so that x=0.

3. Aug 28, 2007

### rocomath

latex!!! i'm kind of confused

however i got the same answer, ~.498

4. Aug 28, 2007

### nophysics

Sorry for the confusion

that x is a multiply sign, I have edited the original post.

I got it now. Thanks.

Last edited: Aug 28, 2007
5. Aug 28, 2007

### rocomath

i did nothing different than plug in what you had typed out into my calculator

$$x=\frac{1}{[1-\exp^{-1.768}]}\times\exp^{-0.884}$$

if you're having trouble typing it into your calculator, this is what i do (for almost everything! and it's so useful for chemistry)

type in the expression then press STO$$\rhd$$ then ALPHA then A then ENTER

let $$A=(1-\exp^{-1.768})$$

let $$B=\exp^{-0.884}$$

$$x=[\frac{B}{A}]$$

Last edited: Aug 28, 2007
6. Aug 28, 2007

### Gib Z

I like doing that when I want to store a computation that I do over a large period of time. For my storage memory A, I have the sum of the harmonic series up to some number. Right now Its up to the 432th term, I add to it whenever I feel bored. I like to see how slow it grows :) And also to see it slowly converging to its asymptotic expansion:

$$\sum_{n=1}^k \frac{1}{n}$$~$$\log_e k + \gamma$$

where $\gamma$ is the Euler-Mascheroni Constant.