If $3 ^ { x } = 4 ^ { x - 1 }$, then $x =$
(A) $\frac { 2 \log _ { 3 } 2 } { 2 \log _ { 3 } 2 - 1 }$
(B) $\frac { 2 } { 2 - \log _ { 2 } 3 }$
(C) $\frac { 1 } { 1 - \log _ { 4 } 3 }$
(D) $\frac { 2 \log _ { 2 } 3 } { 2 \log _ { 2 } 3 - 1 }$

The value of $$\left( \left( \log _ { 2 } 9 \right) ^ { 2 } \right) ^ { \frac { 1 } { \log _ { 2 } \left( \log _ { 2 } 9 \right) } } \times ( \sqrt { 7 } ) ^ { \frac { 1 } { \log _ { 4 } 7 } }$$ is $\_\_\_\_$.

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right hand limit
$$\lim _ { x \rightarrow 0 ^ { + } } \frac { ( 1 - x ) ^ { \frac { 1 } { x } } - e ^ { - 1 } } { x ^ { a } }$$
is equal to a nonzero real number, is $\_\_\_\_$

The product of all positive real values of $x$ satisfying the equation
$$x ^ { \left( 16 \left( \log _ { 5 } x \right) ^ { 3 } - 68 \log _ { 5 } x \right) } = 5 ^ { - 16 }$$
is $\_\_\_\_$.

Let $a = 3 \sqrt { 2 }$ and $b = \frac { 1 } { 5 ^ { 1 / 6 } \sqrt { 6 } }$. If $x , y \in \mathbb { R }$ are such that
$$\begin{aligned} & 3 x + 2 y = \log _ { a } ( 18 ) ^ { \frac { 5 } { 4 } } \\ & 2 x - y = \log _ { b } ( \sqrt { 1080 } ) \end{aligned}$$
then $4 x + 5 y$ is equal to $\_\_\_\_$ .

If the sum of the first 20 terms of the series $\log_{(7^{1/2})}x + \log_{(7^{1/3})}x + \log_{(7^{1/4})}x + \ldots$ is 460, then $x$ is equal to:
(1) $7^2$
(2) $7^{1/2}$
(3) $e^2$
(4) $7^{46/21}$

$\lim _ { x \rightarrow 0 } \left( \tan \left( \frac { \pi } { 4 } + x \right) \right) ^ { 1 / x }$ is equal to
(1) $e$
(2) 2
(3) 1
(4) $e ^ { 2 }$

The sum of the roots of the equation, $x + 1 - 2 \log _ { 2 } 3 + 2 ^ { x } + 2 \log _ { 4 } 10 - 2 ^ { - x } = 0$, is :
(1) $\log _ { 2 } 14$
(2) $\log _ { 2 } 12$
(3) $\log _ { 2 } 13$
(4) $\log _ { 2 } 11$

If for $x \in \left( 0 , \frac { \pi } { 2 } \right) , \log _ { 10 } \sin x + \log _ { 10 } \cos x = - 1$ and $\log _ { 10 } ( \sin x + \cos x ) = \frac { 1 } { 2 } \left( \log _ { 10 } n - 1 \right) , n > 0$, then the value of $n$ is equal to :
(1) 20
(2) 12
(3) 9
(4) 16

If sum of the first 21 terms of the series $\log _ { 9^{1/2} } x + \log _ { 9^{1/3} } x + \log _ { 9^{1/4} } x + \ldots$ where $x > 0$ is 504, then $x$ is equal to
(1) 243
(2) 9
(3) 7
(4) 81

The inverse of $y = 5 ^ { \log x }$ is:
(1) $x = 5 ^ { \log y }$
(2) $x = y ^ { \log 5 }$
(3) $y = x ^ { \frac { 1 } { \log 5 } }$
(4) $x = 5 ^ { \frac { 1 } { \log y } }$

The number of solutions of the equation $\log _ { 4 } ( x - 1 ) = \log _ { 2 } ( x - 3 )$ is

The number of solutions of the equation $\log _ { ( x + 1 ) } \left( 2 x ^ { 2 } + 7 x + 5 \right) + \log _ { ( 2 x + 5 ) } ( x + 1 ) ^ { 2 } - 4 = 0 , x > 0$, is

If $\alpha , \beta$ are the roots of the equation $x ^ { 2 } - \left( 5 + 3 ^ { \sqrt { \log _ { 3 } 5 } } - 5 ^ { \sqrt { \log _ { 5 } 3 } } \right) x + 3 \left( 3 ^ { \left( \log _ { 3 } 5 \right) ^ { \frac { 1 } { 3 } } } - 5 ^ { \left( \log _ { 5 } 3 \right) ^ { \frac { 2 } { 3 } } } - 1 \right) = 0$ then the equation, whose roots are $\alpha + \frac { 1 } { \beta }$ and $\beta + \frac { 1 } { \alpha }$,
(1) $3 x ^ { 2 } - 20 x - 12 = 0$
(2) $3 x ^ { 2 } - 10 x - 4 = 0$
(3) $3 x ^ { 2 } - 10 x + 2 = 0$
(4) $3 x ^ { 2 } - 20 x + 16 = 0$

Let $\beta = \lim _ { x \rightarrow 0 } \frac { \alpha x - \left( e ^ { 3 x } - 1 \right) } { \alpha x \left( e ^ { 3 x } - 1 \right) }$ for some $\alpha \in \mathbb { R }$. Then the value of $\alpha + \beta$ is:
(1) $\frac { 14 } { 5 }$
(2) $\frac { 3 } { 2 }$
(3) $\frac { 5 } { 2 }$
(4) $\frac { 1 } { 2 }$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a function defined by $f ( x ) = \log _ { \sqrt { m } } \{ \sqrt { 2 } ( \sin x - \cos x ) + m - 2 \}$, for some $m$, such that the range of $f$ is $[ 0,2 ]$. Then the value of $m$ is $\_\_\_\_$.
(1) 5
(2) 3
(3) 2
(4) 4

The domain of $f ( x ) = \frac { \log _ { ( x + 1 ) } ( x - 2 ) } { e ^ { 2 \log _ { e } x } - ( 2 x + 3 ) } , x \in R$ is
(1) $\mathbb { R } - \{ - 1,3 \}$
(2) $( 2 , \infty ) - \{ 3 \}$
(3) $( - 1 , \infty ) - \{ 3 \}$
(4) $\mathbb { R } - \{ 3 \}$

Let $a, b, c$ be three distinct positive real numbers such that $2a^{\log_e a} = bc^{\log_e b}$ and $b^{\log_e 2} = a^{\log_e c}$. Then $6a + 5bc$ is equal to $\_\_\_\_$.

The product of all solutions of the equation $\mathrm { e } ^ { 5 \left( \log _ { \mathrm { e } } x \right) ^ { 2 } + 3 } = x ^ { 8 } , x > 0$, is:
(1) $e ^ { 8 / 5 }$
(2) $e ^ { 6 / 5 }$
(3) $e ^ { 2 }$
(4) e

If the domain of the function $\log _ { 5 } \left( 18 x - x ^ { 2 } - 77 \right)$ is $( \alpha , \beta )$ and the domain of the function $\log _ { ( x - 1 ) } \left( \frac { 2 x ^ { 2 } + 3 x - 2 } { x ^ { 2 } - 3 x - 4 } \right)$ is $( \gamma , \delta )$, then $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ is equal to :
(1) 195
(2) 179
(3) 186
(4) 174

Given a sequence $\left\{ a _ { n } \right\}$ that satisfies the following conditions
$$\begin{aligned} & a _ { 1 } = 1 \\ & a _ { n + 1 } = 2 a _ { n } ^ { 2 } \quad ( n = 1,2,3 , \cdots ) , \end{aligned}$$
we are to find the number of natural numbers $n$ satisfying $a _ { n } < 10 ^ { 60 }$. (For the value of $\log _ { 10 } 2$, use the approximation 0.301.)
In this sequence we note that $a _ { n } > 0$ for all natural numbers $n$. Thus when we consider common logarithms of both sides of (1), we have
$$\log _ { 10 } a _ { n + 1 } = \log _ { 10 } \mathbf { A } + \mathbf { B } \log _ { 10 } a _ { n } .$$
When we set $b _ { n } = \log _ { 10 } a _ { n } + \log _ { 10 } \mathbf{A}$, the sequence $\left\{ b _ { n } \right\}$ is a geometric progression such that the common ratio is $\mathbf { C }$. Then
$$\log _ { 10 } a _ { n } = \left( ( \mathbf { D } ) ^ { n - 1 } - \mathbf { E } \right) \log _ { 10 } \mathbf { F } .$$
Furthermore, since $a _ { n } < 10 ^ { 60 }$,
$$\mathbf{D}^{ n - 1 } < \frac { \mathbf { G H } } { \log _ { 10 } \mathbf { F } } + \mathbf { E }$$
Since $\mathbf{IJK}$ is the least natural number which is larger than the value of the right side of (2), the number of natural numbers $n$ satisfying $a _ { n } < 10 ^ { 60 }$ is $\mathbf{L}$.

Consider the following two equations
$$\begin{gathered} \left( \log _ { 4 } 2 \sqrt { x } \right) ^ { 2 } + \left( \log _ { 4 } 2 \sqrt { y } \right) ^ { 2 } = \log _ { 2 } ( \sqrt [ 4 ] { 2 } \cdot x \sqrt { y } ) \\ \sqrt [ 3 ] { x } \cdot \sqrt [ 4 ] { y } = 2 ^ { k } \end{gathered}$$
We are to find the range of values which the constant $k$ can take so that there exist positive real numbers $x , y$ which satisfy (1) and (2) simultaneously.
Set $X = \log _ { 2 } x$ and $Y = \log _ { 2 } y$. Let us express (1) and (2) in terms of $X$ and $Y$. First we consider (1). Since
$$\log _ { 4 } 2 \sqrt { x } = \frac { \log _ { 2 } x + \mathbf { A } } { \mathbf { B } }$$
and
$$\log _ { 2 } ( \sqrt [ 4 ] { 2 } \cdot x \sqrt { y } ) = \frac { \mathbf { C } } { \mathbf { D } } + \log _ { 2 } x + \frac { \log _ { 2 } y } { \mathbf { E } } ,$$
(1) reduces to
$$( X - \mathbf { F } ) ^ { 2 } + ( Y - \mathbf { G } ) ^ { 2 } = \mathbf { H I } .$$
In the same way, (2) reduces to
$$4 X + \mathbf { J } Y = \mathbf { K } \mathbf { L } k .$$
Since the distance $d$ from the center of the circle (3) to the straight line (4) on the $XY$-plane is given by
$$d = \frac { | \mathbf { M N } - \mathbf { O P } k | } { \mathbf { Q } } ,$$
the range of values which $k$ can take is
$$\mathbf { R } \leq k \leqq \mathbf { S } .$$

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 $p > 1$ and $q > 1$. Consider an equation in $x$
$$e ^ { 2 x } - a e ^ { x } + b = 0 \tag{1}$$
such that the equation in $t$ obtained by setting $t = e ^ { x }$ in (1)
$$t ^ { 2 } - a t + b = 0 \tag{2}$$
has the solutions $\log _ { q ^ { 2 } } p$ and $\log _ { p ^ { 3 } } q$. We are to find the minimum value of $a$ and the solution of equation (1) at this minimum.
(1) First of all, we see that
$$b = \frac { \mathbf { A } } { \mathbf { A B } }$$
and
$$a = \frac { \mathbf { C } } { \mathbf{D} } \log _ { q } p + \frac { \mathbf { E } } { \mathbf { F } } \log _ { p } q .$$
(2) As long as $p > 1$ and $q > 1$, it always follows that $\log _ { p } q > \mathbf { G }$. Hence, $a$ takes the minimum value $\frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$ when $\log _ { p } q = \frac { \sqrt { \mathbf { J } } } { \mathbf { K } }$. In this case, the solution of (1) is
$$x = - \frac { \mathbf { L } } { \mathbf { M } } \log _ { e } \mathbf { N } .$$

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 $\mathbf { M }$.