We are given the following information in the image:

$\log_7 2 = a \quad \text{and} \quad \log_2 3 = b$ We need to find $\log_6 98$ in terms of $a$ and $b$.

Solution:

1. Expressing $\log_6 98$ using logarithmic properties: $\log_6 98 = \frac{\log 98}{\log 6}$

2. Breaking down $\log 98$: $98 = 2 \times 49 = 2 \times 7^2$ Therefore, $\log 98 = \log (2 \times 7^2) = \log 2 + 2 \log 7$

3. Breaking down $\log 6$: $6 = 2 \times 3$ Therefore, $\log 6 = \log (2 \times 3) = \log 2 + \log 3$

4. Substituting known values for $\log_7 2 = a$ and $\log_2 3 = b$:

   From the given information:

   \quad \text{and} \quad \log_2 3 = b \implies \log 3 = b \log 2$$ Using these in the equations for $$\log 98$$ and $$\log 6$$: $$\log 98 = \log 2 + 2 \cdot \frac{\log 2}{a}$$ $$\log 6 = \log 2 + b \log 2 = \log 2 (1 + b)$$

5. Final expression for $\log_6 98$: $\log_6 98 = \frac{\log 2 + 2 \cdot \frac{\log 2}{a}}{\log 2 (1 + b)}$

   Factor out $\log 2$ in both the numerator and the denominator: $\log_6 98 = \frac{\log 2 \left( 1 + \frac{2}{a} \right)}{\log 2 (1 + b)} = \frac{1 + \frac{2}{a}}{1 + b}$

Thus, the answer is: $\boxed{ \frac{a + 2}{a(b + 1)} }$

So the correct answer is option C.

Would you like more details or have any further questions? Here are 5 related questions to deepen your understanding:

1. What are the logarithmic properties used to simplify $\log(2 \times 7^2)$?
2. How does the change of base formula work for logarithms?
3. How can you derive logarithms in terms of different bases from a given logarithmic relationship?
4. What would $\log_5 100$ be if you are given $\log_5 2$ and $\log_5 5$?
5. How can logarithms help simplify exponential equations?

Tip: Always simplify logarithmic expressions by breaking them down into known components using logarithmic identities.