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

Question

Find the value of $ \log_2 32 $.

Answer 1

Here is a second way, applying the general rule $\log_{a^m} b^n = \dfrac{n}{m}\log_a b$ directly.

First term. $\log_{27}9 = \log_{3^3} 3^2 = \dfrac{2}{3}\log_3 3 = \dfrac{2}{3}$.
Second term. $\log_{16}64 = \log_{2^4} 2^6 = \dfrac{6}{4}\log_2 2 = \dfrac{3}{2}$.
Denominator term. $\log_4\sqrt2 = \log_{2^2} 2^{1/2} = \dfrac{1/2}{2}\log_2 2 = \dfrac{1}{4}$.
Combine. Numerator $= \dfrac{2}{3}\times\dfrac{3}{2} = 1$. Dividing by the denominator: $\dfrac{1}{1/4} = 4$.
Decimal cross-check. $\log_{27}9 \approx 0.6667$, $\log_{16}64 \approx 1.5$, product $\approx 1.0$. $\log_4\sqrt2 = 0.25$. $1.0/0.25 = 4$, matching exactly.

Both methods confirm the value is 4, so option D is correct. \[ \boxed{4} \]

Answer 2

Here is a second way, applying the general rule $\log_{a^m} b^n = \dfrac{n}{m}\log_a b$ directly.

First term. $\log_{27}9 = \log_{3^3} 3^2 = \dfrac{2}{3}\log_3 3 = \dfrac{2}{3}$.
Second term. $\log_{16}64 = \log_{2^4} 2^6 = \dfrac{6}{4}\log_2 2 = \dfrac{3}{2}$.
Denominator term. $\log_4\sqrt2 = \log_{2^2} 2^{1/2} = \dfrac{1/2}{2}\log_2 2 = \dfrac{1}{4}$.
Combine. Numerator $= \dfrac{2}{3}\times\dfrac{3}{2} = 1$. Dividing by the denominator: $\dfrac{1}{1/4} = 4$.
Decimal cross-check. $\log_{27}9 \approx 0.6667$, $\log_{16}64 \approx 1.5$, product $\approx 1.0$. $\log_4\sqrt2 = 0.25$. $1.0/0.25 = 4$, matching exactly.

Both methods confirm the value is 4, so option D is correct. \[ \boxed{4} \]

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