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skbhatta49 · Jul 26, 2011, 11:15 AM

2^x=x^2
2 to the power X equals x to the power 2
I have found a solution i.e 2 instantly but I found many answers so I operated by log which confused me

ebaines · Jul 26, 2011, 12:20 PM

I'm not sure how you found "many answers" - I believe there are 3.

For positive values of x you can do this:

$ x^2 = 2^ x \\ 2 \ln x = x \ln 2 \\ \frac 1 x \ln x = \frac 1 2 \ln 2\\ \frac 1 x \ln x = \ln ( \sqrt 2) $

If you plot ${f(x) = (1/x) \ln x}$ you'll see that f(x) equals $\ln (\sqrt 2)$ at only two points, those being x = 2 and x=4. To prove that there are the only two solutions - if you find the derivative of this function you'll see that it equals zero only at x = e, so there can be at most two points where it equals $\sqrt 2$ . Hence there are only two solutions for positive values of x.

For negative values of x we can let y = -x, and then a similar approach yields

$ \frac 1 y \ln y = - \ln(\sqrt 2) $

which has a solution at around y = 0.7667, so x = -0.7667 is the third solution. As a check - below is a plot of $y_1 = x^2$ and $y_2 = 2^x$ and you can see the three points where the functions cross.