Question

What is the value of \(\dfrac{\log_{27}9 \times \log_{16}64}{\log_4\sqrt2}\)?

• A. \(\dfrac{1}{6}\)
• B. \(\dfrac{1}{4}\)
• C. 8
• D. 4

Hint:

Convert each log to a simple fraction using powers of the same base first.

Answer 1

The Correct Option is D

Step 1: Simplify each logarithm by writing both numbers as powers of the same base.
\(\log_{27}9\): since \(27=3^3\) and \(9=3^2\), \(\log_{27}9=\dfrac{2}{3}\).
\(\log_{16}64\): since \(16=2^4\) and \(64=2^6\), \(\log_{16}64=\dfrac{6}{4}=\dfrac{3}{2}\).

Step 2: Multiply the numerator terms.
\[ \log_{27}9 \times \log_{16}64 = \frac{2}{3}\times\frac{3}{2} = 1 \]

Step 3: Simplify the denominator.
\(\log_4\sqrt2\): since \(4=2^2\) and \(\sqrt2=2^{1/2}\), \(\log_4\sqrt2=\dfrac{1/2}{2}=\dfrac{1}{4}\).

Step 4: Divide.
\[ \frac{1}{1/4} = 4 \]

Final Answer:
The expression evaluates to 4, so option D is correct. \[ \boxed{4} \]