1. The problem asks to evaluate $\frac{\log_k 32}{\log_k 2}$ for $k > 0$. 2. We use the change of base formula: $\frac{\log_k a}{\log_k b} = \log_b a$. 3. Applying this, we get $\frac{\log_k 32}{\log_k 2} = \log_2 32$. 4. Since $32 = 2^5$, then $\log_2 32 = 5$. 5. Therefore, the value of the expression is $5$. 6. The correct answer is (c) 5.