How do I go about this: X^2 = 2^X?

x1 = -0.7667

x2 = 2

x3 = 4

X=2

Really the easiest way to solve this is to use a graphing calculator.

Plug in y=x^2 and y=2^x and find the intersections.

x= 2, 4, and -.7666647

You can then substitute these answers back into the original equation to check.

There is another way to solve it algebraically, but it's really complicated.

To find all the solutions this required special function called Lambert W-Function, which is a bit complicated.

But the integer solutions are: 2 & 4.

In fact, x = 2 or 4. (integer numbers)

lnx^2 = ln2^x

2lnx = xln2

2/ln2 = x/lnx

I suppose when you get to here, you can see that 2 is definitely one of the answers. But is it the only answer? I guess you can argue that when x increases, the numerator, x, increases linearly, while the denominator, lnx, increases logarithmically, so they are not changing proportionally, and so they will not be equal to the constant 2/ln2 at other values besides x=2.

Hmm, M is correct. In fact, you can see here: http://www.wolframalpha.com/input/?i=y%3Dx%2Flnx+a... that x/lnx = 2/ln2 has two solutions, but only two.

x=2

2^2=2^2...........-_-