1. **State the problem:** Write the expression $3 \log_a(9y + 1) + \frac{1}{4} \log_a(y + 9)$ as a single logarithm. 2. **Recall the logarithm power rule:** For any logarithm, $c \log_a(b) = \log_a(b^c)$. 3. **Apply the power rule to each term:** $$3 \log_a(9y + 1) = \log_a((9y + 1)^3)$$ $$\frac{1}{4} \log_a(y + 9) = \log_a((y + 9)^{\frac{1}{4}})$$ 4. **Use the logarithm addition rule:** $\log_a(M) + \log_a(N) = \log_a(M \cdot N)$. 5. **Combine the two logarithms:** $$\log_a((9y + 1)^3) + \log_a((y + 9)^{\frac{1}{4}}) = \log_a\left((9y + 1)^3 (y + 9)^{\frac{1}{4}}\right)$$ **Final answer:** $$\boxed{\log_a\left((9y + 1)^3 (y + 9)^{\frac{1}{4}}\right)}$$ This is the expression written as a single logarithm.