A solução da equação $e^{2x} = e^5$ é
(A) $x = 1$ (B) $x = \dfrac{5}{2}$ (C) $x = 3$ (D) $x = 4$ (E) $x = 5$

In a certain region, the average number of earthquakes $N$ with magnitude $M$ or greater occurring in one year satisfies the following equation. $$\log N = a - 0.9 M \text{ (where } a \text{ is a positive constant)}$$ In this region, earthquakes with magnitude 4 or greater occur on average 64 times per year. Earthquakes with magnitude $x$ or greater occur on average once per year. Find the value of $9 x$. (Use $\log 2 = 0.3$ for the calculation.) [4 points]

What is the sum of all real roots of the exponential equation $2 ^ { x } + 2 ^ { 2 - x } = 5$? [3 points]
(1) - 2
(2) - 1
(3) 0
(4) 1
(5) 2

Solve the equation $3^{x-8} = \left(\frac{1}{27}\right)^x$ for the real number $x$. [3 points]

The sum of all the solutions of $2 + \log _ { 2 } ( x - 2 ) = \log _ { ( x - 2 ) } 8$ in the interval $( 2 , \infty )$ is
(A) $\frac { 35 } { 8 }$.
(B) 5 .
(C) $\frac { 49 } { 8 }$.
(D) $\frac { 55 } { 8 }$.

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 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 $\_\_\_\_$.

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) $\log _ { 2 } 11$

The number of real roots of the equation $e ^ { 4 x } + 2 e ^ { 3 x } - e ^ { x } - 6 = 0$ is :
(1) 0
(2) 1
(3) 4
(4) 2

The sum of all real roots of equation $\left( e ^ { 2 x } - 4 \right) \left( 6 e ^ { 2 x } - 5 e ^ { x } + 1 \right) = 0$ is
(1) $\ln 4$
(2) $- \ln 3$
(3) $\ln 3$
(4) $\ln 5$

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

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
$$x = - \frac { \mathbf { L } } { \mathbf { M } } \log _ { e } \mathbf { N } .$$

We are to find the minimum value of the function
$$f ( x ) = 8 ^ { x } + 8 ^ { - x } - 3 \left( 4 ^ { 1 + x } + 4 ^ { 1 - x } - 2 ^ { 4 + x } - 2 ^ { 4 - x } \right) - 24$$
and the value of $x$ at which the function takes this minimum value.
First, let us set $2 ^ { x } + 2 ^ { - x } = t$. Then, since
$$4 ^ { x } + 4 ^ { - x } = t ^ { 2 } - \mathbf { A } \quad \text { and } \quad 8 ^ { x } + 8 ^ { - x } = t ^ { 3 } - \mathbf { B } t ,$$
$f ( x )$ can be expressed as
$$f ( x ) = t ^ { 3 } - \mathbf { C D } t ^ { 2 } + \mathbf { E F } t$$
When we consider the right side as a function of $t$ and denote it by $g ( t )$, its derivative is
$$g ^ { \prime } ( t ) = \mathbf { G } ( t - \mathbf { H } ) ( t - \mathbf { I } ) \text {, }$$
where $\mathrm { H } < \mathrm { I }$. Here, since $2 ^ { x } + 2 ^ { - x } = t$, the range of the values which $t$ takes is
$$t \geqq \mathbf { J }$$
When $t = \mathbf { J }$, we see that $g ( \mathbf { J } ) = \mathbf { K L }$. When $t > \mathbf { J }$, $g ( t )$ is locally maximized at $t = \mathbf { M }$, and its local maximum is $\mathbf { N O }$, and furthermore, it is locally minimized at $t = \mathbf { P }$, and its local minimum is $\mathbf { Q R }$.
Thus, the minimum value of $f ( x )$ is $\mathbf { S T }$, which is taken at
$$x = \mathbf { U } \quad \text { and } \quad x = \log _ { 2 } ( \mathbf { V } \pm \sqrt { \mathbf { W X } } ) - \mathbf { Y } .$$

$$2 ^ { 2 x } - 2 \cdot 2 ^ { x } - 8 = 0$$
Given this equation, which of the following is x?
A) 2
B) 1
C) $\ln 2$
D) $\ln 4$
E) $2 \ln 4$

$$4 ^ { x } \cdot 6 ^ { x } \cdot 9 ^ { x } = 36$$
Given this, what is x?
A) $\frac { 2 } { 3 }$
B) $\frac { 1 } { 4 }$
C) $\frac { 3 } { 4 }$
D) $\frac { 3 } { 8 }$
E) $\frac { 4 } { 9 }$

$$\log _ { 8 } \left( \log _ { 9 } ( \sqrt { x + 1 } ) \right) = \frac { - 2 } { 3 }$$
Given this, what is x?
A) 2
B) 3
C) 5
D) 7
E) 8

$$9 ^ { x + 1 } + 3 ^ { x + 1 } - 6 = 0$$
Given this, which of the following is x?
A) $\frac { \ln 3 } { \ln 2 }$
B) $\frac { 1 + \ln 3 } { \ln 2 }$
C) $\frac { 2 + \ln 3 } { \ln 2 }$
D) $\frac { 3 + \ln 2 } { \ln 3 }$
E) $\frac { \ln 2 - \ln 3 } { \ln 3 }$

$$3 ^ { x } \cdot 12 ^ { 2 - x } = 18$$
Given this, what is $x$?
A) $\frac { 1 } { 2 }$
B) $\frac { 3 } { 2 }$
C) $\frac { 1 } { 3 }$
D) $\frac { 4 } { 3 }$
E) $\frac { 5 } { 4 }$

$x ^ { \ln 4 } - 6 \cdot 2 ^ { \ln x } + 8 = 0$\ What is the product of the $x$ values that satisfy this equation?\ A) $e ^ { 6 }$\ B) $e ^ { 4 }$\ C) $e ^ { 3 }$\ D) $\frac { e ^ { 2 } } { 2 }$\ E) $\frac { e ^ { 3 } } { 3 }$