x^2 = 2^x (1 Viewer)

[onebytwo]

does any have any idea how to solve for x.
the only thing i can think of is taking the natural log of both sides, but from then i get stuck
thanks

i know that two solutions are 2 and 4 but i cant work out the third one

 

[A l]

Solve graphically using intersections of y = x² and y = 2^x?

 

[Riviet]

You could apply Newton's method to make an accurate approximation. By trial and error on the calculator, I got a value of about -0.77.

 

[Riviet]

Well I got it off a random e-mail from a friend, thought it was interesting.

 

[Mill]

theres some sort of theorem on this

i believe there are no other solutions ?

u can show that 2^n > n^2 for n > 4 etcz by induction ETCCZZZZ

 

[alcalder]

If you graph it, there is a third solution with -1 < x < -0.75

Newton's method seems the best way to get there. Then show 2^n>n^2 for all n>4 and 2^n<n^2 for all n<-1.

 

[Affinity]

IF x > 0
the equation is same as:
2*ln(x) = x*ln(2)

so if you think a bit, would have up to 2 solutions.

now x=2 or 4 are solutions hence only solutions.

correction..
if x is negative, that equation becomes (where y = -x)
(-y)^2 = 2^(-y)
y^2 = 1/2^y

2ln(y) = -y*ln(2)

then you use numerical methods to get y