What is the value of `(log_(27)9xxlog_(16)64)/(log_(4)sqrt(2))` ?

Options

A. A)`1/6`
B. B)`1/4`
C. C)8
D. D)4

Answer

To solve the expression \((\log_{27} 9 \times \log_{16} 64) / \log_{4} \sqrt{2}\), we will follow these steps:

### Step 1: Rewrite the logarithms using the change of base formula
We can use the change of base formula, which states that \(\log_b a = \frac{\log_k a}{\log_k b}\) for any base \(k\). We will use base 2 for our calculations.

\[ \log_{27} 9 = \frac{\log_2 9}{\log_2 27}, \quad \log_{16} 64 = \frac{\log_2 64}{\log_2 16}, \quad \log_{4} \sqrt{2} = \frac{\log_2 \sqrt{2}}{\log_2 4} \]

### Step 2: Substitute the values into the expression
Now we substitute these values into the original expression:

\[ \frac{\log_{27} 9 \times \log_{16} 64}{\log_{4} \sqrt{2}} = \frac{\frac{\log_2 9}{\log_2 27} \times \frac{\log_2 64}{\log_2 16}}{\frac{\log_2 \sqrt{2}}{\log_2 4}} \]

### Step 3: Simplify the expression
This can be simplified to:

\[ = \frac{\log_2 9 \times \log_2 64 \times \log_2 4}{\log_2 27 \times \log_2 16 \times \log_2 \sqrt{2}} \]

### Step 4: Express logarithms in terms of powers of 2 and 3
Now we express the logarithms in terms of powers:

- \(9 = 3^2\) so \(\log_2 9 = 2 \log_2 3\)
- \(27 = 3^3\) so \(\log_2 27 = 3 \log_2 3\)
- \(64 = 2^6\) so \(\log_2 64 = 6\)
- \(16 = 2^4\) so \(\log_2 16 = 4\)
- \(\sqrt{2} = 2^{1/2}\) so \(\log_2 \sqrt{2} = \frac{1}{2}\)
- \(4 = 2^2\) so \(\log_2 4 = 2\)

### Step 5: Substitute these values back into the expression
Now substituting these values gives:

\[ = \frac{(2 \log_2 3) \times 6 \times 2}{(3 \log_2 3) \times 4 \times \frac{1}{2}} \]

### Step 6: Simplify the expression
This simplifies to:

\[ = \frac{24 \log_2 3}{6 \log_2 3} = \frac{24}{6} = 4 \]

### Final Answer
Thus, the value of the expression is:

\[ \boxed{4} \]