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

[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 xAnswer: 0.498

 

[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.

 

[rocomath]

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.

latex! I'm kind of confused

however i got the same answer, ~.498

 

[nophysics]

Sorry for the confusion

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

I got it now. Thanks.

 

[rocomath]

that x is a multiply sign, I should have left it out.

Roco - how did you do it ? I am curious since I think I messed up at the start.. not sure how to ...

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)

in your calculator (if you don't know already), store the following

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}

so for your answer, you would type in

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

 

[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.