1. **State the problem:** Find the value of $\log_3(b) \times \log_b(27)$. 2. **Recall the change of base formula:** For any positive numbers $a$, $b$, and $c$ (with $a \neq 1$, $b \neq 1$), $$\log_a(c) = \frac{\log_b(c)}{\log_b(a)}$$ 3. **Rewrite $\log_b(27)$ using base 3:** Since 27 is a power of 3, $27 = 3^3$, so $$\log_b(27) = \log_b(3^3) = 3 \log_b(3)$$ 4. **Express $\log_b(3)$ in terms of $\log_3(b)$:** Using the change of base formula, $$\log_b(3) = \frac{1}{\log_3(b)}$$ 5. **Substitute back into the expression:** $$\log_3(b) \times \log_b(27) = \log_3(b) \times 3 \log_b(3) = \log_3(b) \times 3 \times \frac{1}{\log_3(b)}$$ 6. **Simplify by canceling $\log_3(b)$:** $$= 3 \times \cancel{\log_3(b)} \times \frac{1}{\cancel{\log_3(b)}} = 3$$ **Final answer:** $$\log_3(b) \times \log_b(27) = 3$$