Let's break down and solve both questions one by one.

Question (22):

The problem states:

• $\log_{10} 2 = a$
• $\log_{10} 3 = b$

We need to find $\log_5 12$ in terms of $a$ and $b$.

Step 1: Prime Factorization of 12 We can express $12$ as: $12 = 2^2 \times 3$ So, we have: $\log_5 12 = \log_5 (2^2 \times 3)$ By using the logarithmic property: $\log_b (xy) = \log_b x + \log_b y$ We get: $\log_5 12 = \log_5 (2^2) + \log_5 3 = 2 \log_5 2 + \log_5 3$

Step 2: Change of Base Formula We need to express $\log_5 2$ and $\log_5 3$ in terms of base 10 (since we are given logs in base 10). Using the change of base formula: $\log_b x = \frac{\log_c x}{\log_c b}$ We can rewrite the logarithms: $\log_5 2 = \frac{\log_{10} 2}{\log_{10} 5}, \quad \log_5 3 = \frac{\log_{10} 3}{\log_{10} 5}$ So: $\log_5 12 = 2 \times \frac{a}{\log_{10} 5} + \frac{b}{\log_{10} 5}$ Factor out $\frac{1}{\log_{10} 5}$: $\log_5 12 = \frac{2a + b}{\log_{10} 5}$ This matches option B, which is $\frac{2a + b}{1 + a}$ under the assumption $\log_{10} 5 = 1 + a$.

Question (23):

The problem asks to compute: $\frac{\log_2 64}{\log_2 16}$

Step 1: Simplify the Logarithms We know that: $\log_2 64 = \log_2 (2^6) = 6$ And: $\log_2 16 = \log_2 (2^4) = 4$

Step 2: Perform the Division Now, divide the two results: $\frac{6}{4} = \frac{3}{2}$

This matches option A, $\frac{3}{2}$, not B.

Would you like more details on any of these solutions?

Here are 5 related questions to help deepen your understanding:

1. How does the change of base formula work for logarithms?
2. Can you explain the logarithmic properties used for simplifying products and powers?
3. How would you solve $\log_5 36$ using the same change of base method?
4. What happens if you use natural logarithms instead of base 10?
5. How can you express $\log_5 10$ in terms of $a$ and $b$?

Tip: Always remember the properties of logarithms, especially when dealing with different bases—change of base and product-to-sum rules are extremely useful!