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May 4th, 2008, 01:21 PM   #1
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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.
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May 5th, 2008, 07:57 AM   #2
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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!
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May 5th, 2008, 08:37 AM   #3
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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
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May 5th, 2008, 10:30 AM   #4
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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!
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May 5th, 2008, 06:07 PM   #5
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Please tell me if you know of one. I've really tried everything and I'm running out of ideas
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May 7th, 2008, 02:41 PM   #6
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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.
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May 7th, 2008, 04:39 PM   #7
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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?
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May 8th, 2008, 12:15 AM   #8
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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.
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May 8th, 2008, 05:41 AM   #9
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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.
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May 8th, 2008, 06:18 AM   #10
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Quote:
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.
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