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

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 20,475 Thanks: 2039 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)? $-2^{e \cdot x}-x+2=0$ $-2^{e \cdot x}=x-2 /\cdot e$ $-e\cdot 2^{e \cdot x}=e\cdot x-2e$ $-e\cdot 2^{e \cdot x-2e}\cdot 2^{2e}=e\cdot x-2e$ $e\cdot 2^{e \cdot x-2e}\cdot 2^{2e}=2e - e\cdot x$ $e\cdot 2^{2e}=\left(2e - e\cdot x\right) \cdot 2^{2e - e \cdot x}$ $e\cdot 2^{2e}=\left(2e - e\cdot x\right) \cdot e^{\left(2e - e \cdot x\right)\ln 2} / \cdot \ln2$ $e\cdot 2^{2e}\cdot \ln 2=\left(2e - e\cdot x\right) \cdot \ln 2\cdot e^{\left(2e - e \cdot x\right)\ln 2}$ $\left(2e - e\cdot x\right) \cdot \ln 2= W\left(e\cdot 2^{2e}\cdot \ln 2\right) / \cdot \frac{1}{e \cdot \ln 2}$ $2 - x= \frac{1}{e\cdot\ln 2}W\left(e\cdot 2^{2e}\cdot \ln 2\right)$ $x=2- \frac{1}{e\cdot\ln 2}W\left(e\cdot 2^{2e}\cdot \ln 2\right)$

 Tags advanced, algebraically, difficult, extremely, solve

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post forkconfig Applied Math 1 February 6th, 2014 02:25 PM GrannySmith Algebra 2 January 5th, 2014 03:50 PM drewm Algebra 2 July 4th, 2011 03:06 PM problem_solver123 Algebra 2 September 5th, 2009 09:37 PM Zan Geometry 2 May 23rd, 2008 07:22 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top