Algebra question: x^2 = 2^x?

Question

Solve x for x^2 = 2^x

This is a specific instance of the x^y = y^x problem which is transcendental, and explained in greater detail in another Yahoo answers and also Dr. Math. There are multiple solutions, and I don't know if I found all of them, hence this question.

Answer 1

Well, two solutions are easy....
One of them is obvious: x=2
The second was found by someone already x=4
Are there more? Yes.
We have to consider the function
f(x)=2^x-x^2
If it is zero for a given x, the x solves our equation.
For x=0, the function is 1
For x-> -inf, f goes to -inf
so in between there must be a solution.
if you get to it numerically, you get something like -0.766665
That about wraps it up, I guess.

Answer 2

x = 2 or x = 4

Answer 3

the answers are 2, 4 and appx. -.77

try graphing y = x^2 -2^x
and note points where graph crosses x-axis

fyi 0 is not a solution:

0^2 = 0 and 2^0 = 1

and neither is -2
-2^2 = 4 and 2^-2 = 1/4

Answer 4

use your distributive resources in both horizontal or vertical formats: x^2(x^2-2x+a million) +2x(x^2-2x+a million)-a million(x^2-2x+a million) and ruin each and each and every grouping.... till you are able to combine like words in descending exponentail order

Answer 5

also, x=y, or x=2, y=2
and perhaps
x=2
y=-2

x could also equal 0 if I'm not mistaken.

x could definitely be infinity, but I don't think that is a rational equation.

Maybe x is an imaginary number in some instances, like the square root of negative 1.

Answer 6

I can think of this example.


2^4 = 16 and 4^2 = 16

Answer 7

If we take ln of both sides
2ln IxI =xln2
So take f(x) = 2 ln IxI-x ln 2
At 0 lim f(x) = -infinity
lim f(x) x=>- infinity = +infinity
lim f(x) x=>+infinity = - infinity as in x(2lnx/x-ln2) lnx/x has limit 0
f´(x)= 2/x-ln2=0 x=2/ln2
the sign of f´(x) is

--------- 0+++++++2/ln2---------
so at x=2/ln2 we have a local maximum
f(2/ln2) =0.119 >0 so there are two positive roots
There is one negative root between -1 and 0 as at x=-1 f(-1) >0 and at 0 lin f(x)=- infinity
f(-0.5) =-1.039<0 so the root is between -1 and -0.5