The Logistic model is one of the commonly used mathematical models and can be applied in epidemiology. Based on published data, scholars established a Logistic model for the cumulative confirmed cases of COVID-19 in a certain region $I ( t )$ (where $t$ is measured in days): $I ( t ) = \frac { K } { 1 + \mathrm { e } ^ { -0.23 ( t - 53 ) } }$, where $K$ is the maximum number of confirmed cases. When $I \left( t ^ { * } \right) = 0.95 K$, it indicates that the epidemic has been initially controlled. Then $t ^ { * }$ is approximately (given $\ln 19 \approx 3$)
A. 60
B. 63
C. 66
D. 69

The Logistic model is one of the commonly used mathematical models and can be applied in epidemiology. Based on published data, scholars established a Logistic model for the cumulative confirmed cases $I ( t )$ of COVID-19 in a certain region ($t$ in days): $I ( t ) = \frac { K } { 1 + \mathrm { e } ^ { - 0.23 ( t - 53 ) } }$ , where $K$ is the maximum number of confirmed cases. When $I \left( t ^ { * } \right) = 0.95 K$ , it indicates that the epidemic has been initially controlled. Then $t ^ { * }$ is approximately ( $\ln 19 \approx 3$ )
A. 60
B. 63
C. 66
D. 69

If real numbers $x, y, z$ satisfy $2 + \log_2 x = 3 + \log_3 y = 5 + \log_5 z$, then the size relationship of $x, y, z$ that is impossible is
A. $x > y > z$
B. $x > z > y$
C. $y > x > z$
D. $y > z > x$

If real numbers $x, y, z$ satisfy $2 + \log_2 x = 3 + \log_3 y = 5 + \log_5 z$, then the size relationship of $x, y, z$ that is impossible is
A. $x > y > z$
B. $x > z > y$
C. $y > x > z$
D. $y > z > x$

Find all values of $c$ for which the equation $\log_2 x = cx$ has solutions.

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 $$\lim _ { x \rightarrow 0 } \left[ 1 + x \ln \left( 1 + b ^ { 2 } \right) \right] ^ { \frac { 1 } { x } } = 2 b \sin ^ { 2 } \theta , b > 0 \text { and } \theta \in ( - \pi , \pi ]$$ then the value of $\theta$ is
(A) $\pm \frac { \pi } { 4 }$
(B) $\pm \frac { \pi } { 3 }$
(C) $\pm \frac { \pi } { 6 }$
(D) $\pm \frac { \pi } { 2 }$

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

Consider the function $y = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } }$, where $x \geqq 0$.
(1) We are to find the $x$ at which $y$ is minimized.
When we differentiate $y$, we have
$$\frac { d y } { d x } = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } \left( 2 x \log _ { e } \mathbf { A } - \mathbf { B } \log _ { e } \mathbf { C } \right) .$$
Hence, when we express the value of $x$ at which $y$ is minimized using the common logarithm, we have
$$x = \frac { \mathbf { D } } { \mathbf { F } \left( 1 - \log _ { 10 } \mathbf { E } \right) } .$$
(2) We are to find the smallest positive integer $x$ satisfying $\frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } > 1000$.
From the inequality $y > 1000$, we obtain
$$x ^ { \mathbf { H } } \log _ { 10 } \mathbf { I } - \mathbf { J } x \log _ { 10 } \mathbf { L } - \mathbf { K } > 0 .$$
When we solve the inequality using $0.3$ as an approximate value for $\log _ { 10 } 2 = 0.301 \cdots$, the smallest positive integer $x$ satisfying $y > 1000$ is $\mathbf{Q}$.

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
$$x = \mathbf { U } \quad \text { and } \quad x = \log _ { 2 } ( \mathbf { V } \pm \sqrt { \mathbf { W X } } ) - \mathbf { Y } .$$

In the early stages of an infectious disease outbreak, since most people have not been infected and have no antibodies, the total number of infected people usually grows exponentially. Under the premise that ``the initial number of infected people is $P _ { 0 }$ , and each infected person on average infects $r$ uninfected people per day'', the total number of people infected with the disease after $n$ days, $P _ { n }$ , can be expressed as
$$P _ { n } = P _ { 0 } ( 1 + r ) ^ { n } \text {, where } P _ { 0 } \geq 1 \text { and } r > 0 \text { . }$$
Answer the following questions:
(1) Given that $A = \frac { \log P _ { 5 } - \log P _ { 2 } } { 3 } , B = \frac { \log P _ { 8 } - \log P _ { 6 } } { 2 }$ , show that $A = B$ . (4 points)
(2) Given that a certain infectious disease in its early stages follows the above mathematical model and the total number of infected people increases tenfold every 16 days, find the value of $\frac { P _ { 20 } } { P _ { 17 } } \times \frac { P _ { 8 } } { P _ { 6 } } \times \frac { P _ { 5 } } { P _ { 2 } }$ . (5 points)
(3) Based on (2), find the value of $\frac { \log P _ { 20 } - \log P _ { 17 } } { 3 }$ . (4 points)

According to experimental statistics, a certain type of bacteria reproduces such that its quantity increases by a factor of 2.4 on average every 3.5 hours. Suppose a test tube in the laboratory initially contains 1000 of this type of bacteria. According to an exponential function model, approximately how many hours later will the quantity of this bacteria reach about $4 \times 10^{10}$? (Note: $\log 2 \approx 0.3010$, $\log 3 \approx 0.4771$)
(1) 63 hours
(2) 70 hours
(3) 77 hours
(4) 84 hours
(5) 91 hours

Let $x _ { 0 }$、$y _ { 0 }$ be positive real numbers. If the point $\left( 10 x _ { 0 } , 100 y _ { 0 } \right)$ on the coordinate plane lies on the graph of the function $y = 10 ^ { x }$ , then the point $\left( x _ { 0 } , \log y _ { 0 } \right)$ will lie on the graph of the line $y = a x + b$ , where $a$、$b$ are real numbers. What is the value of $2 a - b$?
(1) 4
(2) 9
(3) 15
(4) 18
(5) 22

If $x, y$ are two positive real numbers satisfying $x^{-\frac{1}{3}} y^{2} = 1$ and $2\log y = 1$, then $\frac{x - y^{2}}{10} =$ (13--1) (13--2).

The half-life $T$ of a radioactive substance is defined as ``every time period $T$ passes, the mass of the substance decays to half of its original amount''. A lead container contains two radioactive substances $A$ and $B$ with half-lives $T _ { A }$ and $T _ { B }$ respectively. At the start of recording, the masses of these two substances are equal. After 112 days, measurement shows that the mass of substance $B$ is one-quarter of the mass of substance $A$. Based on the above, which of the following is the relationship between $T _ { A }$ and $T _ { B }$?
(1) $- 2 + \frac { 112 } { T _ { A } } = \frac { 112 } { T _ { B } }$
(2) $2 + \frac { 112 } { T _ { A } } = \frac { 112 } { T _ { B } }$
(3) $- 2 + \log _ { 2 } \frac { 112 } { T _ { A } } = \log _ { 2 } \frac { 112 } { T _ { B } }$
(4) $2 + \log _ { 2 } \frac { 112 } { T _ { A } } = \log _ { 2 } \frac { 112 } { T _ { B } }$
(5) $2 \log _ { 2 } \frac { 112 } { T _ { A } } = \log _ { 2 } \frac { 112 } { T _ { B } }$

A student derived an equation that two physical quantities $s$ and $t$ should satisfy. To verify the theory, he conducted an experiment and obtained 15 sets of data for the two physical quantities $(s_{k}, t_{k})$, $k = 1, \cdots, 15$. The teacher suggested that he first take the logarithm of $t_{k}$, and plot the corresponding points $\left(s_{k}, \log t_{k}\right)$, $k = 1, \cdots, 15$ on the coordinate plane; where the first data is the horizontal axis coordinate and the second data is the vertical axis coordinate. Using regression line analysis, the student verified his theory. The regression line passes through the origin with a positive slope less than 1. What is the relationship between $s$ and $t$ that the student obtained most likely to be which of the following options?
(1) $s = 2t$
(2) $s = 3t$
(3) $t = 10^{s}$
(4) $t^{2} = 10^{s}$
(5) $t^{3} = 10^{s}$

Research shows that the residual amount of a certain drug in a user's body decreases exponentially over time after taking the drug. It is known that 2 hours after taking the drug, half of the drug dose remains in the body. Which of the following options is correct?
(1) After 3 hours, the body still retains $\frac{1}{3}$ of the drug dose
(2) After 4 hours, the body still retains $\frac{1}{4}$ of the drug dose
(3) After 6 hours, the body still retains $\frac{1}{6}$ of the drug dose
(4) After 8 hours, the body still retains $\frac{1}{8}$ of the drug dose
(5) After 10 hours, the body still retains $\frac{1}{10}$ of the drug dose

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

$$\frac { 3 ^ { x } } { 2 ^ { 2 x } } = \frac { 1 } { 5 }$$
Given this, what is the value of the expression $5 ^ { \frac { 1 } { x } }$?
A) $\frac { 3 } { 2 }$
B) $\frac { 4 } { 3 }$
C) $\frac { 9 } { 4 }$
D) $\frac { 9 } { 5 }$
E) $\frac { 5 } { 6 }$