1. ## 3-x=3e^(-x)

hello everyone, I am currently working on an assignment and have solved my problem down to the following:

3-x=3e^(-x)

I have graphed both functions and seen that a solution exists that is not 0 and I am lost as to how to find the exact solution.

If anyone can help that would be greatly appreciated.

thanks,
Eric

2. Originally Posted by Eric.Physics
hello everyone, I am currently working on an assignment and have solved my problem down to the following:

3-x=3e^(-x)

I have graphed both functions and seen that a solution exists that is not 0 and I am lost as to how to find the exact solution.

If anyone can help that would be greatly appreciated.

thanks,
Eric
you will not be able to find an exact solution using elementary algebraic methods ... x = 0 can be determined by observation ... the other solution can be found using numerical methods (technolgy).

3. I was hoping that wasn't the case. Oh well, thanks for your help

4. This isn't what skeeter (or I) would call an "elementary algebraic method" but:
Multiply both sides by $\displaystyle e^x$: $\displaystyle (3- x)e^x= 3$ or $\displaystyle (x- 3)e^x= -3$. Now let y= x- 3 so that x= y- 3 and the equation becomes $\displaystyle ye^{y-3}= e^{-3}[ye^y]= -3$ so $\displaystyle ye^y= -3e^3$.

Now use "Lambert's W function" which is defined as the inverse to the function f(x)= $\displaystyle xe^x$. In this case $\displaystyle y= x- 3= W(-3e^3)$ so that $\displaystyle x= W(-3e^3)+ 3$. That is, as I said, not an "elementary algebraic" method but it is as good a solution as "x= cos(3)" or "x= log(5)" would be.