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...