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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!

If log10 2 = a, log10 3 = b, then log5 12 = ?; log2 64 / log2 16 = ?

Learn how to solve logarithmic equations involving log10 2 = a, log10 3 = b, and find log5 12 using the change of base formula. Also, compare logarithms with log2 64 / log2 16. Suitable for students in Grades 9-10.

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