For the function with domain $\{ x \mid 1 \leqq x \leqq 81 \}$, $$y = \left( \log _ { 3 } x \right) \left( \log _ { \frac { 1 } { 3 } } x \right) + 2 \log _ { 3 } x + 10$$ Let $M$ be the maximum value and $m$ be the minimum value. Find the value of $M + m$. [4 points]

What is the minimum value of the function $y = 3 + \log _ { 3 } \left( x ^ { 2 } - 4 x + 31 \right)$? [3 points]
(1) 4
(2) 5
(3) 6
(4) 7
(5) 8

12. Given $a > 0 , b > 0 , ab = 8$, then when $a$ equals $\_\_\_\_$, $\log _ { 2 } a \cdot \log _ { 2 } ( 2 b )$ attains its maximum value.

Let $m$ be the minimum possible value of $\log _ { 3 } \left( 3 ^ { y _ { 1 } } + 3 ^ { y _ { 2 } } + 3 ^ { y _ { 3 } } \right)$, where $y _ { 1 } , y _ { 2 } , y _ { 3 }$ are real numbers for which $y _ { 1 } + y _ { 2 } + y _ { 3 } = 9$. Let $M$ be the maximum possible value of ( $\log _ { 3 } x _ { 1 } + \log _ { 3 } x _ { 2 } + \log _ { 3 } x _ { 3 }$ ), where $x _ { 1 } , x _ { 2 } , x _ { 3 }$ are positive real numbers for which $x _ { 1 } + x _ { 2 } + x _ { 3 } = 9$. Then the value of $\log _ { 2 } \left( m ^ { 3 } \right) + \log _ { 3 } \left( M ^ { 2 } \right)$ is

Let $x$ and $y$ be positive numbers which satisfy
$$\left(\log_2 x\right)^2 + \left(\log_2 y\right)^2 = \log_2 \frac{8x^2}{y^2}. \tag{1}$$
We are to find the maximum value of $xy^2$ and the values of $x$ and $y$ at that point.
(1) The right side of (1) can be transformed into
$$\log_2 \frac{8x^2}{y^2} = \mathbf{A}\log_2 x - \mathbf{B}\log_2 y + \mathbf{C}.$$
So, setting $X = \log_2 x$ and $Y = \log_2 y$, we can express (1) in terms of $X$ and $Y$ as
$$(X - \mathbf{D})^2 + (Y + \mathbf{E})^2 = \mathbf{F}.$$
(2) Set $k = \log_2 xy^2$. Using $X$ and $Y$ above, this equality can be transformed into
$$X + \mathbf{GG}\,Y - k = 0.$$
If we graph (2) and (3) on a plane with coordinates $(X, Y)$, the graph of (2) is a circle, and the graph of (3) is a straight line. When $k$ is maximized, the graph of (3) is tangent to the graph of (2). Hence, when $k = \mathbf{H}$, $xy^2$ takes the maximum value IJ and in this case $x = \mathbf{K}$ and $y = \mathbf{L}$.

Let $x$ satisfy the inequality
$$2 \left( \log _ { \frac { 1 } { 3 } } x \right) ^ { 2 } + 9 \log _ { \frac { 1 } { 3 } } x + 9 \leqq 0 .$$
We are to find the maximum value of the function
$$f ( x ) = \left( \log _ { 3 } x \right) \left( \log _ { 3 } \frac { x } { 3 } \right) \left( \log _ { 3 } \frac { x } { 9 } \right) .$$
The range of values of $x$ satisfying (1) is
$$\mathbf { A } \sqrt { \mathbf { B } } \leqq x \leqq \mathbf { C D } .$$
When we set $t = \log _ { 3 } x$, the range of values of $t$ is
$$\frac { \mathbf { E } } { \mathbf { F } } \leqq t \leqq \mathbf { G } .$$
When we express the right side of (2) in terms of $t$ and consider it as a function $g ( t )$, its derivative is
$$g ^ { \prime } ( t ) = \mathbf { H } t ^ { 2 } - \mathbf { I } t + \mathbf { J } .$$
Hence $f ( x )$ is maximized at $x = \mathbf { K L }$, and its maximum value is $\mathbf { M }$.