   SEARCH HOME Math Central Quandaries & Queries  This has been bugging me for ages. How do find both real solutions to x^x = 2x? Obviously I have x=2, and there's another at about 0.35, but I can't work it out properly. Any help? Hi Ramsay,

I don't know any way to arrive at an algebraic form for the second solution. The only technique I know is to approximate it.

I don't know how you arrived at an approximation of 0.35, but once you have an approximation you can continue to improve it. For example if you let

f(x) = xx - 2x

then you are looking for x that satisfies f(x) = 0.

f(0.35) = -0.00749 and f(0.34) = 0.0129 and thus the solution is somewhere between 0.34 and 0.35.

Then I calculated f(0.345) = 0.00270 so the solution is between 0.345 and 0.35.

You can divide this interval in half to further improve the approximation and continue.

There are other approximation methods, the best known is the Newton-Raphson method, but an approximation is the best you can do.

Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.