We are asked to evaluate the expression $\log_{9} \sqrt[6]{32}(\log_{2} 3 + \log_{4} 9 + \log_{8} 27 + \log_{16} 81)$.

1. Problem Description

We are asked to evaluate the expression $\log_{9} \sqrt[6]{32}(\log_{2} 3 + \log_{4} 9 + \log_{8} 27 + \log_{16} 81)$.

2. Solution Steps

First, let's simplify $\log_{2} 3 + \log_{4} 9 + \log_{8} 27 + \log_{16} 81$.

Using the change of base formula, $\log_{a^n} b^m = \frac{m}{n} \log_{a} b$, we have:

$\log_{2} 3 = \log_{2} 3$

$\log_{4} 9 = \log_{2^2} 3^2 = \frac{2}{2} \log_{2} 3 = \log_{2} 3$

$\log_{8} 27 = \log_{2^3} 3^3 = \frac{3}{3} \log_{2} 3 = \log_{2} 3$

$\log_{16} 81 = \log_{2^4} 3^4 = \frac{4}{4} \log_{2} 3 = \log_{2} 3$

Therefore, $\log_{2} 3 + \log_{4} 9 + \log_{8} 27 + \log_{16} 81 = 4 \log_{2} 3$.

Now, we can rewrite the original expression as:

$\log_{9} \sqrt[6]{32} (4 \log_{2} 3) = \log_{9} (32^{\frac{1}{6}} (4 \log_{2} 3))$.

We know that $32 = 2^5$, so $32^{\frac{1}{6}} = (2^5)^{\frac{1}{6}} = 2^{\frac{5}{6}}$.

Thus, $\log_{9} (2^{\frac{5}{6}} \cdot 4 \log_{2} 3) = \log_{9} (2^{\frac{5}{6}} \cdot 2^2 \log_{2} 3) = \log_{9} (2^{\frac{17}{6}} \log_{2} 3)$.

This seems difficult to simplify further. Let's rewrite the expression in terms of a common base.

Original expression: $\log_{9} \sqrt[6]{32} (\log_{2} 3 + \log_{4} 9 + \log_{8} 27 + \log_{16} 81) = \log_{9} (32^{\frac{1}{6}}) (4 \log_{2} 3)$

$= \log_{3^2} (2^5)^{\frac{1}{6}} (4 \log_{2} 3) = \log_{3^2} (2^{\frac{5}{6}} \cdot 2^2 \log_{2} 3) = \log_{3^2} (2^{\frac{17}{6}} \log_{2} 3)$

$= \frac{1}{2} \log_{3} (2^{\frac{17}{6}} \log_{2} 3)$

Using the property $\log_{a} (xy) = \log_{a} x + \log_{a} y$,

$= \frac{1}{2} (\log_{3} 2^{\frac{17}{6}} + \log_{3} (\log_{2} 3))$

$= \frac{1}{2} (\frac{17}{6} \log_{3} 2 + \log_{3} (\log_{2} 3))$

$= \frac{17}{12} \log_{3} 2 + \frac{1}{2} \log_{3} (\log_{2} 3)$.

This looks complicated. Let's go back to the original expression.

$\log_{9} \sqrt[6]{32} (4 \log_{2} 3) = \log_{9} \sqrt[6]{2^5} (4 \log_{2} 3) = \log_{3^2} 2^{\frac{5}{6}} (4 \log_{2} 3) = \frac{1}{2} \log_{3} (2^{\frac{5}{6}} \cdot 4 \log_{2} 3)$

$= \frac{1}{2} \log_{3} (2^{\frac{5}{6}}) + \frac{1}{2} \log_{3} (4 \log_{2} 3) = \frac{1}{2} \cdot \frac{5}{6} \log_{3} 2 + \frac{1}{2} \log_{3} (4 \log_{2} 3)$

$= \frac{5}{12} \log_{3} 2 + \frac{1}{2} \log_{3} 4 + \frac{1}{2} \log_{3} (\log_{2} 3) = \frac{5}{12} \log_{3} 2 + \frac{1}{2} \log_{3} 2^2 + \frac{1}{2} \log_{3} (\log_{2} 3)$

$= \frac{5}{12} \log_{3} 2 + \log_{3} 2 + \frac{1}{2} \log_{3} (\log_{2} 3) = \frac{17}{12} \log_{3} 2 + \frac{1}{2} \log_{3} (\log_{2} 3)$.

Let us use change of base to base 2

$4 \log_2 3 = \log_2 3 + \log_2 3 + \log_2 3 + \log_2 3 = \log_2 3 + \log_4 9 + \log_8 27 + \log_{16} 81$

So we want to evaluate $\log_9 \sqrt[6]{32}(4\log_2 3)$

Using the change of base formula $\log_b a = \frac{\log_c a}{\log_c b}$.

Then $\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2}$.

$\log_9 \sqrt[6]{32} (4\log_2 3) = \log_9 (2^{\frac{5}{6}}) (4\log_2 3) = \frac{1}{2}\log_3 (2^{\frac{5}{6}} 4\log_2 3)$

If $\log_2 3 = x$, then we have

$\frac{1}{2}\log_3 (2^{\frac{5}{6}} 4x) = \frac{1}{2}\log_3 (2^{\frac{5}{6} + 2} x) = \frac{1}{2}\log_3 (2^{\frac{17}{6}} x)$

$= \frac{1}{2}[\frac{17}{6} \log_3 2 + \log_3 x ] = \frac{17}{12}\log_3 2 + \frac{1}{2} \log_3 \log_2 3$

Let's start with:

$\log_9 \sqrt[6]{32} (4\log_2 3) = \log_9 \sqrt[6]{2^5} (4\log_2 3) = \log_9 2^{\frac{5}{6}} + \log_9 4\log_2 3 = \frac{5}{6} \log_9 2 + \log_9 4 + \log_9 \log_2 3 = \frac{5}{6} \log_{3^2} 2 + \log_{3^2} 4 + \log_{3^2} \log_2 3$

$= \frac{5}{12} \log_3 2 + \frac{1}{2} \log_3 4 + \frac{1}{2} \log_3 \log_2 3 = \frac{5}{12} \log_3 2 + \log_3 2 + \frac{1}{2} \log_3 \log_2 3 = \frac{17}{12} \log_3 2 + \frac{1}{2} \log_3 \log_2 3$.

Try converting back to original base 2:

$\log_2 3 = y$. $\log_9 (\sqrt[6]{32}) (4y) = \log_{3^2} 2^{\frac{5}{6}} 4y = \frac{1}{2} [\log_3 (2^{\frac{5}{6}}) + \log_3 (4y)] = \frac{1}{2} [\frac{5}{6}\log_3 2 + \log_3 4 + \log_3 y] = \frac{1}{2} [\frac{5}{6} \frac{1}{\log_2 3} + 2\log_3 2 + \log_3 y] = \frac{1}{2} [\frac{5}{6} \frac{1}{y} + \frac{2}{y} + \log_3 y ]$

$= \frac{1}{2} [\frac{5}{6} \frac{1}{\log_2 3} + 2 \frac{1}{\log_2 3} + \log_3 \log_2 3] = \frac{17}{12}\log_3 2 + \frac{1}{2}\log_3 \log_2 3$.

$\sqrt[6]{32} = 32^{\frac{1}{6}} = (2^5)^{\frac{1}{6}} = 2^{\frac{5}{6}}$.

$\log_9 (\sqrt[6]{32}) = \log_9 2^{\frac{5}{6}} = \frac{5}{6} \log_9 2 = \frac{5}{6} \frac{\log_2 2}{\log_2 9} = \frac{5}{6} \frac{1}{2\log_2 3} = \frac{5}{12} \frac{1}{\log_2 3}$

$\log_9 (\sqrt[6]{32}(4 \log_2 3)) = \log_9 \sqrt[6]{32} + \log_9 4 \log_2 3 = \frac{5}{6} \log_9 2 + \log_9 (4 \log_2 3)$.

$= \frac{5}{6} \frac{\log_2 2}{\log_2 9} + \frac{\log_2 4 \log_2 3}{\log_2 9} = \frac{5}{6} \frac{1}{2\log_2 3} + \frac{\log_2 4 + \log_2 (\log_2 3)}{2\log_2 3}$

$= \frac{5}{12\log_2 3} + \frac{2 + \log_2(\log_2 3)}{2\log_2 3} = \frac{5}{12 \log_2 3} + \frac{1}{\log_2 3} + \frac{1}{2\log_2 3} \log_2(\log_2 3) = \frac{17}{12 \log_2 3} + \frac{1}{2} \frac{\log_2 (\log_2 3)}{\log_2 3}$.

$\log_{2} 3 > 1$.

We have $\log_9 (\sqrt[6]{32} \cdot (4\log_2 3)$.

Let $x = \log_2 3$, then we have $\log_9 (2^{\frac{5}{6}} \cdot 4x)$. Then $\frac{\log_3 2^{\frac{5}{6}} \cdot 4x}{\log_3 9} = \frac{1}{2} \log_3 (2^{\frac{5}{6}} \cdot 4x) = \frac{1}{2} (\frac{5}{6} \log_3 2 + \log_3 4 + \log_3 x) = \frac{1}{2}(\frac{5}{6} \log_3 2 + 2\log_3 2 + \log_3 x) = \frac{1}{2} (\frac{17}{6} \log_3 2 + \log_3 x) = \frac{17}{12} \log_3 2 + \frac{1}{2} \log_3 \log_2 3$.

If $\log_2 3 = 1.58$, so $\frac{17}{12} (.63) + \frac{1}{2} (-.273) \approx \frac{17}{12}(.63) - 0.13 = 0.8925 - 0.13 = 0.7625$.

The options are $\frac{5}{3} = 1.67$, $\frac{3}{2} = 1.5$, $\frac{3}{4} = 0.75$, $1$.

Final Solution: The final answer is $\boxed{1}$

3. Final Answer

(D) 1