Let ` log_(2) 3 = alpha `, then ` log_(64) 729` is

Options

A. `alpha`
B. `(alpha)/(3)`
C. `3 alpha`
D. `2 alpha `

Answer

To solve the problem, we need to find the value of \( \log_{64} 729 \) given that \( \log_{2} 3 = \alpha \).

### Step-by-step Solution:

1. **Rewrite the logarithm using the change of base formula:**
\[ \log_{64} 729 = \frac{\log 729}{\log 64} \]

2. **Express 729 and 64 in terms of their prime factors:**
- \( 729 = 3^6 \)
- \( 64 = 2^6 \)

3. **Substitute these values into the logarithm:**
\[ \log_{64} 729 = \frac{\log(3^6)}{\log(2^6)} \]

4. **Use the power rule of logarithms:**
\[ \log(3^6) = 6 \log 3 \quad \text{and} \quad \log(2^6) = 6 \log 2 \]
Therefore,
\[ \log_{64} 729 = \frac{6 \log 3}{6 \log 2} \]

5. **Simplify the expression:**
The \( 6 \) in the numerator and denominator cancels out:
\[ \log_{64} 729 = \frac{\log 3}{\log 2} \]

6. **Substitute the value of \( \log 3 / \log 2 \) using the given information:**
Since \( \log_{2} 3 = \alpha \), we have:
\[ \frac{\log 3}{\log 2} = \alpha \]

7. **Final answer:**
Thus, we conclude that:
\[ \log_{64} 729 = \alpha \]