The value of `((log)_2 24)/((log)_(96)2)-((log)_2 192)/((log)_(12)2)` is 3 (b) 0 (c) 2 (d) 1

Options

A. 3
B. 0
C. 2
D. 1

Explanation

To solve the expression \(\frac{\log_2 24}{\log_{96} 2} - \frac{\log_2 192}{\log_{12} 2}\), we will convert all logarithms to base 2 using the change of base formula.

### Step-by-step Solution:

1. **Convert the logarithms using the change of base formula**:
\[ \log_{b} a = \frac{\log_k a}{\log_k b} \]
Hence, we can rewrite the expression as:
\[ \frac{\log_2 24}{\log_{96} 2} = \frac{\log_2 24}{\frac{1}{\log_2 96}} = \log_2 24 \cdot \log_2 96 \]
and
\[ \frac{\log_2 192}{\log_{12} 2} = \frac{\log_2 192}{\frac{1}{\log_2 12}} = \log_2 192 \cdot \log_2 12 \]
Thus, the expression becomes:
\[ \log_2 24 \cdot \log_2 96 - \log_2 192 \cdot \log_2 12 \]

2. **Express the numbers in terms of logarithms**:
- We can express \(24\), \(96\), \(192\), and \(12\) in terms of their prime factors:
- \(24 = 2^3 \cdot 3\)
- \(96 = 2^5 \cdot 3\)
- \(192 = 2^6 \cdot 3\)
- \(12 = 2^2 \cdot 3\)

3. **Calculate the logarithms**:
- Using the property \(\log_b (mn) = \log_b m + \log_b n\):
\[ \log_2 24 = \log_2 (2^3 \cdot 3) = 3 + \log_2 3 \]
\[ \log_2 96 = \log_2 (2^5 \cdot 3) = 5 + \log_2 3 \]
\[ \log_2 192 = \log_2 (2^6 \cdot 3) = 6 + \log_2 3 \]
\[ \log_2 12 = \log_2 (2^2 \cdot 3) = 2 + \log_2 3 \]

4. **Substitute back into the expression**:
\[ = (3 + \log_2 3)(5 + \log_2 3) - (6 + \log_2 3)(2 + \log_2 3) \]

5. **Expand both products**:
\[ = (15 + 8\log_2 3 + \log_2^2 3) - (12 + 8\log_2 3 + \log_2^2 3) \]

6. **Simplify the expression**:
\[ = 15 - 12 + 8\log_2 3 - 8\log_2 3 + \log_2^2 3 - \log_2^2 3 \]
\[ = 3 \]

### Final Answer:
Thus, the value of the expression is \(3\).