The problem in the image reads:

Given: $\log_3 a = x \quad \text{and} \quad \log_a b = y,$ evaluate: $\log_6 b.$

Solution:

We start by expressing $\log_6 b$ in terms of the given variables $x$ and $y$.

Step 1: Change the base of $\log_6 b$ to base $a$:

$\log_6 b = \frac{\log_a b}{\log_a 6}.$

From the given information, we know: $\log_a b = y.$

Thus: $\log_6 b = \frac{y}{\log_a 6}.$

Step 2: Express $\log_a 6$ in terms of $x$:

We know that $6 = 3 \cdot 2$. Using the product rule for logarithms: $\log_a 6 = \log_a 3 + \log_a 2.$

From the first given equation ($\log_3 a = x$), rewrite it as: $\log_a 3 = \frac{1}{x}.$

Substitute this into $\log_a 6$: $\log_a 6 = \frac{1}{x} + \log_a 2.$

Step 3: Express $\log_a 2$ in terms of $\log_3 a$ (if needed):

If $\log_a 2$ is unknown, the problem may require additional context for further simplification. For now, the expression becomes:

$\log_6 b = \frac{y}{\frac{1}{x} + \log_a 2}.$

This is the evaluated expression for $\log_6 b$ in terms of $x$ and $y$. If additional values or relationships (e.g., $\log_a 2$) are known, the solution can be simplified further.

Do you have any clarifications or additional details? Let me know!

Here are 5 follow-up questions based on this problem:

1. How does the change of base formula work in logarithms?
2. Can you derive $\log_a 2$ in terms of $x$ if additional values are provided?
3. What happens if $x = 1$ and $y = 1$? What is $\log_6 b$ then?
4. Why do we use logarithmic properties to simplify such problems?
5. How do the laws of logarithms help solve real-world problems?

Tip:

Always rewrite logarithmic expressions in terms of a single base if possible; this simplifies solving and substitution.