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Solve for $x$ where: $$x^2=2^x$$

Graphing the function ($f(x)=x^2-2^x$) it is clear that there are three solutions.

Two of them are nice solutions. i.e. $x=2,4$

The graph shows a third solution at $x=-0.76666...$

I found this using the Newton-Raphson method but it took quite a few iterations.

What other methods are there for solving such a question?

Kantura
  • 2,721
  • I think he is looking for an alternative numerical method . – Khosrotash Jun 03 '17 at 19:50
  • @Derek:http://web.cecs.pdx.edu/~gerry/class/ME350/pdf/ME350_Lecture_05_slides_root-finding.pdf – Khosrotash Jun 03 '17 at 19:53
  • @derek:there is some popular method to find $f(x)=0$ roots ,such as Fixed point iteration • Bisection • Newton’s method • Secant method – Khosrotash Jun 03 '17 at 19:55
  • 2 by inspection; 4 by guess and check; the third one is goign to be difficult no matter what. – Jacob Wakem Jun 03 '17 at 20:34
  • The negative solution is solution of $x=-e^{\ln2/2·x}$ or $-\ln2/2·x·e^{-\ln2/2·x}=\ln2/2$ so that with the Lambert-W function $\ln2/2·x=-W(\ln2/2)$. With solution $x=-0.76666469596212306$ – Lutz Lehmann Jun 04 '17 at 18:10

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