# Thread: Solve for 'x' algebraically...difficult and need help?

1. ## Solve for 'x' algebraically...difficult and need help?

I need to know how to algebraically solve for 'x' in this equation:

Note: ' * ' denotes multiplication.

-2^(e*x)-x+2=0

Please, show all of your steps because no matter what I do, I wind up going back and fourth.

The answer should be:

x = ((2*e*log(2)-Productlog((2^2e)*e*log(2))/(e*log(2)))

x ≈ 0.285983

I don't even know what "Productlog" is. I've heard of it used in a Lambert Function, but that's for programming and whatnot, so I'm not sure if it's the same. Grrr. If there are any other ways of solving this, please explain.

Any help is appreciated, thanks.

2. Originally Posted by IsenseHelp
I need to know how to algebraically solve for 'x' in this equation:

Note: ' * ' denotes multiplication.

-2^(e*x)-x+2=0

Please, show all of your steps because no matter what I do, I wind up going back and fourth.

The answer should be:

x = ((2*e*log(2)-Productlog((2^2e)*e*log(2))/(e*log(2)))

x ≈ 0.285983

I don't even know what "Productlog" is. I've heard of it used in a Lambert Function, but that's for programming and whatnot, so I'm not sure if it's the same. Grrr. If there are any other ways of solving this, please explain.

Any help is appreciated, thanks.
$\displaystyle \Rightarrow -2 e^x = x - 2$

Substitute $\displaystyle t = x - 2$:

$\displaystyle -2 e^{t+2} = t \Rightarrow -2 e^2 = t e^{-t}$

$\displaystyle \Rightarrow 2 e^2 = -t e^{-t}$

$\displaystyle \Rightarrow -t = W(2 e^2)$

$\displaystyle \Rightarrow x = 2 - W(2 e^2)$

where W is the Lambert W-function.