SOLUTION: Solve for x. sqrt{x}^(x) = x^(sqrt{x}) Crazy question, right?

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Question 1196903: Solve for x.
sqrt{x}^(x) = x^(sqrt{x})
Crazy question, right?

Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
A plot of +y+=+sqrt%28x%29%5Ex+-+x%5Esqrt%28x%29+ shows 1 and 4 as the only likely solutions on the real numbers.
+graph%28400%2C400%2C+-10%2C+10%2C+-10%2C+10%2C+sqrt%28x%29%5Ex+-+x%5Esqrt%28x%29+%29+

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

Your starting equation is 

    sqrt%28x%29%5Ex = x%5E%28sqrt%28x%29%29.       (1)


One solution is x= 1 (obvious).


Indeed,  left side is sqrt%281%29%5E1 = 1%5E1 = 1.

         Right side is equal to 1 due to the same reason.


Now I will assume that x =/= 1 and will look for other solutions.



Take logarithm base 10 of both sides of equation (1).  You will get

    x%2Alog%28%28sqrt%28x%29%29%29 = sqrt%28x%29%2Alog%28%28x%29%29.


Divide both sides by log(x)  (we can do it safely, since we consider  x =/= 1.)
You will get

    sqrt%28x%29%2Fx = log%28%28sqrt%28x%29%29%29%2Flog%28%28x%29%29,


or, equivalently,

    sqrt%28x%29%2Fx = 1%2F2.


It implies

    2%2Asqrt%28x%29 = x

and after squaring both sides,

    4x = x%5E2.


It implies

    4x+-+x%5E2 = 0

    x*(4-x) = 0,

    x = 0  or   x= 4.


You can check that the root x= 0 works in the original equation, since then each side is equal to 1.

So, the remaining solutions to the problem are  x= 0  and  x= 4.


ANSWER.  The given equation has three solutions  x= 0,  x= 1,  and x= 4.

Solved. 

                  To see that x= 0 is the solution, too, look at this plot below.



    +graph%28800%2C400%2C+-0.2%2C+1.2%2C+-1%2C+2%2C+sqrt%28x%29%5Ex%2C+x%5Esqrt%28x%29+%29+


                        Plots y = sqrt%28x%29%5Ex  (red)  and  y = x%5Esqrt%28x%29  (green)