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mycrafish

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

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- Thread starter mycrafish
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- #1

mycrafish

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

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- #3

NascentOxygen

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

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HallsofIvy

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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/webMathematica/FunctionEvaluation.jsp?name=ProductLog. That's what I used to get the value above.)

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

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