Let's break down and solve the given problem step by step.

Part 1: Evaluate the individual logarithms.

1. $\log_3 27$:

   We know that $27 = 3^3$, so: $\log_3 27 = \log_3 (3^3) = 3$ This follows from the logarithmic rule $\log_b (b^x) = x$.

2. $\log_7 \left( \frac{1}{49} \right)$:

   We know that $49 = 7^2$, so $\frac{1}{49} = 7^{-2}$. Therefore: $\log_7 \left( \frac{1}{49} \right) = \log_7 (7^{-2}) = -2$ Again, using the rule $\log_b (b^x) = x$.

Part 2: Solve the equation.

The equation to solve is: $\log_3 27 + \log_7 \left( \frac{1}{49} \right) = \log_9 y$ From the previous calculations, we know: $\log_3 27 = 3 \quad \text{and} \quad \log_7 \left( \frac{1}{49} \right) = -2$ Substitute these values into the equation: $3 + (-2) = \log_9 y$ This simplifies to: $1 = \log_9 y$ Now, solving for $y$, using the property $\log_b a = c \implies a = b^c$: $y = 9^1 = 9$

Final Answer:

• $y = 9$

Would you like any further details or have any questions?

Here are some related questions:

1. What is the change of base formula for logarithms?
2. How do logarithmic properties simplify complex equations?
3. What are the key properties of exponents used in logarithmic equations?
4. How does one solve logarithmic equations with different bases?
5. Can you explain the relationship between logarithms and exponential functions?

Tip: Always remember to express numbers in terms of powers of the base when working with logarithms to simplify the evaluation process.