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 May 4th, 2008, 01:21 PM #1 Newbie   Joined: May 2008 Posts: 9 Thanks: 0 equation Anyone know how to solve a^x = bx? I know you have to put it in Ae^A = k where A = -ln(a)x and k = -ln(a) * b^-1. Then you take Ae^A = k and say A = w(k). This works for one solution, but what about the other solution? Take 4^x = 8x. With the above method you get x = 0.15 which is right. But 2 is also right and how do you find it? I hope someone knows something about this because I really wanna know. Please forgive me if I posted this the wrong place. Thanks  May 5th, 2008, 07:57 AM #2 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Hm... First or all, please put ... on the end of your solution "x=0.15" because it's not correct otherwise.. And regarding your other question, you're trying to find the inverse or f(x) = 4^x - 8x that's not bijective. So, without realizing it, you've limited yourself only to solutions x < 1.26438... That's why you're not able to get x = 2. Hope you get it! May 5th, 2008, 08:37 AM #3 Newbie   Joined: May 2008 Posts: 9 Thanks: 0 Yes, sorry about that. 0.15.... of course! But what did you mean? I can't find 2? If you know a formula, please explain, I'm loosing my mind here  May 5th, 2008, 10:30 AM #4 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 No, here's the deal.. When you've introduced the Lambert W function into your equation, you lost your x = 2 solution. It's quite similar when solving x^2=4. If you were to introduce the function f(x)=sqrt(x) then you'd get x = 2, but lack the solution x = -2. It's the same case here, but more complicated. Yet another example is solving sin x = 0. Using the f(x) = arcsin(x) function you get x = 0, but don't get an infinitive number of solutions that are gained by analyzing the sin function with more care. So, to answer your question - to get the second solution you'd need to use yet another Lambert-W like function.. It's just the trick of finding one!  May 5th, 2008, 06:07 PM #5 Newbie   Joined: May 2008 Posts: 9 Thanks: 0 Please tell me if you know of one. I've really tried everything and I'm running out of ideas  May 7th, 2008, 02:41 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 The Lambert W-function is sometimes called the Product Log function, and is available in Mathematica as ProductLog[k, z], where k is 0 if you want the principal value (also available as ProductLog[z] or LambertW[z]), or -1 if you want the value which gives the second real solution for the above problem. May 7th, 2008, 04:39 PM #7 Newbie   Joined: May 2008 Posts: 9 Thanks: 0 This is great! I feel I understand little by little. But could you please show me how to find 2 in 4^x = 8x for me? I think I could understand if you show me. I don't know much about functions, but I'm thinking maybe the other solution is the inverse of the first or something like that. Also I don't quite understand ProductLog[k, z]. What does the , do? Does it mean it's ProductLog from k to z and how do I use that in my equation? May 8th, 2008, 12:15 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 As you know, 4^x = 8x can be rearranged as Ae^A = -ln(4)/8, where A = -ln(4)x. Hence x = -w(-ln(4)/8)/ln(4), where w is a suitable function. For real values of A, Ae^A is decreasing for A ≤ -1 and increasing for A ≥ -1. For the second range, the inverse is called the principal value of the Lambert W-function and denoted by W; in Mathematica software, W(z) is implemented as LambertW[z] or ProductLog[z], but ProductLog can also accept a second parameter (that is an integer written before z and separated from it by a comma). ProductLog[0, z] is equivalent to ProductLog[z], both giving W(z). For the first range, the inverse function, w, that you need is implemented in Mathematica as ProductLog[-1, z] and ProductLog[-1, -ln(4)/8] has the value -2ln(4). Hence x = -w(-ln(4)/8)/ln(4) simplifies to x = 2. If you don't have the Mathematica software, you can solve Ae^A = z by an iterative successive approximation method of your choice, such as Newton-Raphson. May 8th, 2008, 05:41 AM #9 Newbie   Joined: May 2008 Posts: 9 Thanks: 0 Ok, that was simple But why does W[-1, -ln(4)/8] become -2ln(4)? If it said W[-1, -ln(16)/6], then that would become -8ln(3) or what? Thanks for the great info, and I use the wolfram site to solve W-function so that's no problem. May 8th, 2008, 06:18 AM   #10
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 Originally Posted by skipjack If you don't have the Mathematica software, you can solve Ae^A = z by an iterative successive approximation method of your choice, such as Newton-Raphson.
There are better ways that are as easy to program, or easier. Google "Lambert W function" and "Halley loop" or "Halley iteration" and you should find it. Tags equation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post PhizKid Differential Equations 0 February 24th, 2013 10:30 AM mich89 Algebra 3 January 9th, 2013 01:22 PM xujack88 Calculus 8 December 11th, 2009 07:04 PM

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