Math 110 Course Resources - Exponential \& Logarithmic Functions Course Packet on solving for an unknown exponent Consider the equation \( 2^{t^{2}-24}=(1 / 4)^{t} \). Enter the number of solutions of the above equation. Determine the values of \( t \) that solve the above equation. Enter your answer as a comma-separated list of values. The order of the values does not matter. Enter DNE if there are no solutions.

To solve the equation \( 2^{t^{2}-24} = (1/4)^{t} \), we first recognize that \( 1/4 \) can be rewritten as \( 2^{-2} \). Replacing \( 1/4 \) gives us: \[ 2^{t^{2}-24} = (2^{-2})^{t} = 2^{-2t} \] Since the bases are the same (both are powers of 2), we can set the exponents equal to each other: \[ t^{2} - 24 = -2t \] Rearranging this yields the quadratic equation: \[ t^{2} + 2t - 24 = 0 \] Now we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2, c = -24 \): \[ t = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2} \] Calculating the two possible solutions gives us: 1. \( t = \frac{8}{2} = 4 \) 2. \( t = \frac{-12}{2} = -6 \) Thus, there are two solutions to the original equation: \( t = 4, -6 \). In summary, there are 2 solutions: \( -6, 4 \).