The given problem involves logarithmic expressions and a geometric mean $G$. Let's break it down:

Given:

• $\log_9{x} = \log_{36}{y} = a$
• $G = \sqrt{xy}$
• We need to find $\log_{18}{G}$.

Step 1: Express $x$ and $y$ in terms of $a$

From the given information:

• $\log_9{x} = a$, so $x = 9^a$
• $\log_{36}{y} = a$, so $y = 36^a$

Step 2: Calculate $G = \sqrt{xy}$

Now, substitute $x$ and $y$: $G = \sqrt{xy} = \sqrt{9^a \cdot 36^a}$ This can be rewritten as: $G = \sqrt{(9 \cdot 36)^a} = \sqrt{324^a} = (324^a)^{1/2} = 324^{a/2}$ Thus, $G = 324^{a/2}$.

Step 3: Calculate $\log_{18}{G}$

Now, we need to find $\log_{18}{G}$. We already have $G = 324^{a/2}$, so: $\log_{18}{G} = \log_{18}(324^{a/2})$ Using the logarithmic property $\log_b(a^c) = c \log_b(a)$, this becomes: $\log_{18}{G} = \frac{a}{2} \log_{18}{324}$

Step 4: Simplify $\log_{18}{324}$

Since $324 = 18^2$, we have: $\log_{18}{324} = \log_{18}{(18^2)} = 2$

Step 5: Final Expression

Substitute $\log_{18}{324} = 2$ into the equation: $\log_{18}{G} = \frac{a}{2} \cdot 2 = a$

Thus, $\log_{18}{G} = a$.

Final Answer:

$\log_{18}{G} = a$

Would you like further details or have any other questions?

Follow-up Questions:

1. How do you simplify logarithmic expressions involving powers and products?
2. What other properties of logarithms can help solve similar problems?
3. How does the base of a logarithm affect its value?
4. Can you think of an example where logarithms of different bases are used?
5. How can logarithmic equations be used to model real-world phenomena?

Tip:

When dealing with logarithms, always look for opportunities to simplify using logarithmic properties such as the product, quotient, and power rules.