How do you solve for $x$ when $x^x=2x$? Graphing both $y^x=y+x$ and $y=x$, there are two intersection points, one at $x=2$, and another at $x\approx0.34$. I've recently learned the Lambert-W function, but haven't been able to apply it to this problem.
How to solve for x when $x^x=2x$
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calculus
numerical-methods
lambert-w
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0https://www.wolframalpha.com/input/?i=x%5Ex%3D2x – 2017-01-22
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0The other solution is at $0.346323362278580922$, but apparantly Wolfram does not display a solution involving the Lambert-W-function. Not sure, whether it helps here. – 2017-01-22
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0I would suggest the bisection-method or newton's method to calculate the smaller solution. – 2017-01-22
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0If instead of 2 on the RHS, you had $x^x=Ax$ where $A =e\frac{1}{e}^\frac{1}{e} \approx 1.88$ it would have closed form roots of $1/e, A$. So you're only about 6% away from having an equation with closed form roots :-). – 2017-05-04
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0In fact $\frac{A}{2e} \approx 0.3461$ is a good approximation (within .07%) to the solution – 2017-05-04
1 Answers
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As said in comments, there is no solutions in terms of Lambert function and, then, as already suggested, Newton method is probably the simplest to be used.
Looking for the zero of $$f(x)=x^x-2x$$ $$f'(x)=x^x (\log (x)+1)-2$$ and using $x_0=\frac 12$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.5 \\ 1 & 0.33573216085622873 \\ 2 & 0.34626771116191797 \\ 3 & 0.34632336075977158 \\ 4 & 0.34632336227858092 \end{array} \right)$$