
#1
Mar1712, 07:41 AM

P: 4

x*log(x)=0.1*x^2




#2
Mar1712, 08:07 AM

P: 380

My guess is that using Newton's Method would be the easiest. I don't see any easy way to isolate x in that equation.




#3
Mar1712, 05:57 PM

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P: 4,716





#4
Mar1712, 10:37 PM

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How to figure out answer to logarithm
First, of course, x= 0 is not a solution because log(0) is not defined. So you can divide both sides by x to get log(x)= 0.1 x. You can now write this as [itex]x= e^{0.1x}[/itex]. If you let y= 0.1x, that becomes [itex]y/0.1= e^{y}[/itex]. Now multiply on both sides by [itex]0.1e^y[/itex] to get [itex]ye^y= 0.1[/itex].
Now we can take the Lambert W function of both sides: y= 0.1x= W(0.1) so x= 10W(0.1)= 10(0.111833)= 1.11833 (to six significant figures). (The Lambert W function is defined as the inverse function to [itex]f(x)= xe^x[/itex]. It is also known as the "ProductLog" function. Mathematica evaluates that function and it can be evaluated at http://functions.wolfram.com/webMath...ame=ProductLog. That's what I used to get the value above.) 


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