If `log_(48) 81 = x`, then `log_(12)3` = ______.

Options

A. `(x+4)/(2x)`
B. `(x+4)/(x)`
C. `(x)/(x+4)`
D. `(2x)/(x+4)`

Explanation

To solve the problem, we need to find the value of \( \log_{12} 3 \) given that \( \log_{48} 81 = x \).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \log_{48} 81 = x \]

2. **Express 81 in terms of powers of 3:**
\[ 81 = 3^4 \]
So we can rewrite the logarithm:
\[ \log_{48} (3^4) = x \]

3. **Use the power rule of logarithms:**
According to the power rule, \( \log_b (a^n) = n \cdot \log_b a \):
\[ 4 \cdot \log_{48} 3 = x \]
Therefore, we can express \( \log_{48} 3 \):
\[ \log_{48} 3 = \frac{x}{4} \]

4. **Use the change of base formula:**
The change of base formula states that \( \log_a b = \frac{1}{\log_b a} \):
\[ \log_{48} 3 = \frac{1}{\log_3 48} \]
Hence, we can write:
\[ \frac{x}{4} = \frac{1}{\log_3 48} \]

5. **Rearranging gives us:**
\[ \log_3 48 = \frac{4}{x} \]

6. **Express 48 in terms of its prime factors:**
\[ 48 = 16 \cdot 3 = 4^2 \cdot 3 \]
Thus, we can write:
\[ \log_3 48 = \log_3 (4^2 \cdot 3) = \log_3 (4^2) + \log_3 3 \]

7. **Using the power rule again:**
\[ \log_3 (4^2) = 2 \cdot \log_3 4 \]
Since \( \log_3 3 = 1 \), we have:
\[ \log_3 48 = 2 \cdot \log_3 4 + 1 \]

8. **Equating the two expressions for \( \log_3 48 \):**
\[ 2 \cdot \log_3 4 + 1 = \frac{4}{x} \]

9. **Isolate \( \log_3 4 \):**
\[ 2 \cdot \log_3 4 = \frac{4}{x} - 1 \]
\[ \log_3 4 = \frac{2}{x} - \frac{1}{2} \]

10. **Now, find \( \log_{12} 3 \):**
Since \( 12 = 4 \cdot 3 \):
\[ \log_{12} 3 = \log_{12} (4 \cdot 3) = \log_{12} 4 + \log_{12} 3 \]
Using the change of base formula:
\[ \log_{12} 4 = \frac{\log_3 4}{\log_3 12} \]
And \( \log_3 12 = \log_3 (4 \cdot 3) = \log_3 4 + \log_3 3 = \log_3 4 + 1 \).

11. **Substituting back:**
\[ \log_{12} 3 = \frac{\log_3 4}{\log_3 4 + 1} \]

12. **Substituting \( \log_3 4 \):**
\[ \log_{12} 3 = \frac{\frac{2}{x} - \frac{1}{2}}{\frac{2}{x} - \frac{1}{2} + 1} \]

### Final Answer:
Thus, we have derived the expression for \( \log_{12} 3 \) in terms of \( x \).