d) \( a^{x}+b^{y}=(a b)^{x y} \)

To analyze the equation \( a^{x} + b^{y} = (ab)^{xy} \), we can start by rewriting the right-hand side: \[ (ab)^{xy} = a^{xy} b^{xy} \] Thus, the equation becomes: \[ a^{x} + b^{y} = a^{xy} b^{xy} \] This equation suggests a relationship between the terms \( a^x \), \( b^y \), and \( a^{xy} b^{xy} \). To explore this further, we can consider specific values for \( a \), \( b \), \( x \), and \( y \). ### Case 1: \( a = b = 1 \) If \( a = 1 \) and \( b = 1 \), then: \[ 1^{x} + 1^{y} = 1 + 1 = 2 \] \[ (1 \cdot 1)^{xy} = 1^{xy} = 1 \] This does not satisfy the equation. ### Case 2: \( a = b = 2 \) If \( a = 2 \) and \( b = 2 \), then: \[ 2^{x} + 2^{y} = 2^{x} + 2^{y} \] \[ (2 \cdot 2)^{xy} = 4^{xy} = (2^2)^{xy} = 2^{2xy} \] This is true if \( x = y = 1 \) since: \[ 2^{1} + 2^{1} = 2 + 2 = 4 \] \[ 2^{2 \cdot 1 \cdot 1} = 2^{2} = 4 \] ### General Case To find a more general solution, we can analyze the equation further. 1. **Assume \( x = y \)**: Let \( x = y = k \). Then the equation simplifies to: \[ a^{k} + b^{k} = (ab)^{k^2} \] 2. **Testing with \( k = 1 \)**: \[ a + b = ab \] This implies \( ab - a - b = 0 \) or \( (a-1)(b-1) = 1 \). This gives us pairs \( (a, b) \) such as \( (2, 2) \) or \( (3, 1) \). 3. **Testing with \( k = 2 \)**: \[ a^{2} + b^{2} = (ab)^{4} \] This is a more complex equation and may not yield simple integer solutions. ### Conclusion The equation \( a^{x} + b^{y} = (ab)^{xy} \) holds for specific values of \( a \), \( b \), \( x \), and \( y \). The simplest case is when \( a = b = 2 \) and \( x = y = 1 \). Further exploration may yield additional solutions, but they will likely depend on the specific values chosen for \( a \) and \( b \). If you have specific values for \( a \), \( b \), \( x \), and \( y \) in mind, please provide them for a more tailored analysis!