Solve for \( t \) \( 2^{2 t}-2^{t+2}=32 \)

Let's start by rewriting the equation \( 2^{2t} - 2^{t+2} = 32 \). We can express \( 32 \) as a power of \( 2 \): \( 32 = 2^5 \). Now, we can rewrite the left side of the equation: \( 2^{2t} - 2^{t+2} = 2^{2t} - 2^2 \cdot 2^t = 2^{2t} - 4 \cdot 2^t \). Let’s set \( x = 2^t \). Then, the equation becomes: \( x^2 - 4x = 32 \). Rearranging gives us a standard quadratic equation: \( x^2 - 4x - 32 = 0 \). Now we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 1 \), \( b = -4 \), and \( c = -32 \): \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm \sqrt{144}}{2} = \frac{4 \pm 12}{2}. \] This gives us two possible solutions for \( x \): 1. \( x = \frac{16}{2} = 8 \) 2. \( x = \frac{-8}{2} = -4 \) (not valid since \( x = 2^t \) must be positive) So, we have \( x = 8 \). Now, recall that \( x = 2^t \), thus: \( 2^t = 8 \). Since \( 8 = 2^3 \), we can conclude: \( t = 3 \). Therefore, the solution is: \[ \boxed{3}. \]