If 2^n=n^2, what is the value of n?

 To solve the equation \(2^n = n^2\), we need to find the value(s) of \(n\) that satisfy this condition. 

Let's check some values of \(n\):

1. For \(n = 1\):

   \[

   2^1 = 2 \quad \text{and} \quad 1^2 = 1

   \]

   \(2 \neq 1\)

2. For \(n = 2\):

   \[

   2^2 = 4 \quad \text{and} \quad 2^2 = 4

   \]

   \(4 = 4\) (so \(n = 2\) is a solution)

3. For \(n = 3\):

   \[

   2^3 = 8 \quad \text{and} \quad 3^2 = 9

   \]

   \(8 \neq 9\)

4. For \(n = 4\):

   \[

   2^4 = 16 \quad \text{and} \quad 4^2 = 16

   \]

   \(16 = 16\) (so \(n = 4\) is also a solution)

For values greater than 4, \(2^n\) grows exponentially while \(n^2\) grows polynomially. Thus, \(2^n\) will quickly exceed \(n^2\), and no further integer solutions are found.

Hence, the integer solutions to \(2^n = n^2\) are \(n = 2\) and \(n = 4\).