c. \( 3^{2-x}-3^{-x-3}=\frac{242}{9} \)

To solve the equation \( 3^{2-x}-3^{-x-3}=\frac{242}{9} \), we can start by rewriting both terms in terms of a common base. First, observe that: \[ 3^{2-x} = \frac{9}{3^x} \quad \text{and} \quad 3^{-x-3} = \frac{1}{3^{x+3}} = \frac{1}{27 \cdot 3^x} \] Thus, substituting these in gives: \[ \frac{9}{3^x} - \frac{1}{27 \cdot 3^x} = \frac{242}{9} \] Now, find a common denominator for the left side: \[ \frac{9 \cdot 27 - 1}{27 \cdot 3^x} = \frac{242}{9} \] Calculating \(9 \cdot 27\): \[ 243 - 1 = 242 \] So we have: \[ \frac{242}{27 \cdot 3^x} = \frac{242}{9} \] Equating the fractions yields: \[ 27 \cdot 3^x = 9 \] Dividing both sides by 9 gives: \[ 3 \cdot 3^x = 1 \implies 3^x = \frac{1}{3} \] This leads us to: \[ x = -1 \] Thus, the solution to the equation is: \[ \boxed{-1} \]