My Math Forum Solve for 'x' algebraically, extremely difficult (advanced)?

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 January 7th, 2009, 01:15 PM #1 Newbie   Joined: Jan 2009 Posts: 2 Thanks: 0 Solve for 'x' algebraically, extremely difficult (advanced)? 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 it can't be the same. Grrr. Thanks.
 January 7th, 2009, 03:55 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 Yes, it can be! See this article.
 January 11th, 2009, 04:10 PM #3 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Re: Solve for 'x' algebraically, extremely difficult (advanced)? $-2^{e \cdot x}-x+2=0$ $-2^{e \cdot x}=x-2 /\cdot e$ $-e\cdot 2^{e \cdot x}=e\cdot x-2e$ $-e\cdot 2^{e \cdot x-2e}\cdot 2^{2e}=e\cdot x-2e$ $e\cdot 2^{e \cdot x-2e}\cdot 2^{2e}=2e - e\cdot x$ $e\cdot 2^{2e}=\left(2e - e\cdot x\right) \cdot 2^{2e - e \cdot x}$ $e\cdot 2^{2e}=\left(2e - e\cdot x\right) \cdot e^{\left(2e - e \cdot x\right)\ln 2} / \cdot \ln2$ $e\cdot 2^{2e}\cdot \ln 2=\left(2e - e\cdot x\right) \cdot \ln 2\cdot e^{\left(2e - e \cdot x\right)\ln 2}$ $\left(2e - e\cdot x\right) \cdot \ln 2= W\left(e\cdot 2^{2e}\cdot \ln 2\right) / \cdot \frac{1}{e \cdot \ln 2}$ $2 - x= \frac{1}{e\cdot\ln 2}W\left(e\cdot 2^{2e}\cdot \ln 2\right)$ $x=2- \frac{1}{e\cdot\ln 2}W\left(e\cdot 2^{2e}\cdot \ln 2\right)$

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