Can You Solve x*log(x) = 0.1*x^2 Using Logarithmic Properties?

[mycrafish]

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

 

[scurty]

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.

 

[NascentOxygen]

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

Is log = log₁₀? Then I'd guess about, oh, let's see, ... 10?

 

[HallsofIvy]

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 $x= e^{0.1x}$. If you let y= -0.1x, that becomes $-y/0.1= e^{-y}$. Now multiply on both sides by $-0.1e^y$ to get $ye^y= -0.1$.

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 $f(x)= xe^x$. 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.)