Question 1

Graph the points and from visual inspection, select the model that would best fit the data. Choose from $\begin{array} { l l } 
y = m x + b \text { (linear) } & y = a b ^ { x } \text { (exponential) } \
y = a + b \ln x \text { (logarithmic) } & y = \frac { c } { 1 + a e ^ { - b x } } \text { (logistic) }
\end{array}$ Then use a graphing utility to find a function that fits the data. (Hint: For a logistic model, go to STAT, CALC,
Logistic.)
-$\begin{array} { c | c } 
x & y \
\hline 4 & 4.4 \
5 & 5.2 \
9 & 10.5 \
14 & 25.1 \
20 & 71.3 \
24 & 143.1
\end{array}$ 
A) $y = - 101.2 + 61.88 \ln x$
B) $y = 2.2 ( 1.19 ) ^ { x }$
C) $y = \frac { 425 } { 1 + 214 e ^ { - 0.11 x } }$
D) $y = 6.2 x - 35.27$

Accepted Answer

The answer of Graph the points and from visual inspection,...

Question 2

Solve the problem.
-The population of a country is modeled by the function $P ( t ) = 15.9 e ^ { 0.01462 t }$ where $P ( t )$ is the population (in millions) $t$ years after January 1, 2000. Use the model to predict the year during which year the population will reach 30 million if this trend continues.
A) 2,043
B) 2,014
C) 2,084
D) 2,004

Accepted Answer

To find the year when the population will reach 30 million, set $P(t) = 30$ and solve for $t$:$$30 = 15.9e^{0.01462t}$$Divide both sides by 15.9:$$e^{0.01462t} = \frac{30}{15.9}$$Take the natural logarithm of both sides:$$0.01462t = \ln\left(\frac{30}{15.9}\right)$$Solve for $t$:$$t = \frac{\ln\left(\frac{30}{15.9}\right)}{0.01462}$$$$t \approx 43$$Since $t$ is the number of years after January 1, 2000, the year during which the population will reach 30 million is $2000 + 43 = 2043$.

Question 3

Solve the problem.
-The function $P ( t ) = 8 ( 1.0134 ) ^ { t }$ represents the population (in millions) of a country $t$ years after January 1, 2000. Write an equivalent function using base $e$; that is, write a function of the $P ( t ) = P _ { 0 } e ^ { r t }$. Round $r$ to 5 decimal places. Also, determine the population for the year 2000 .
A) $P ( t ) = 8 e ^ { 0.01331 t }$; population in 2000: 8 million
B) $P ( t ) = 8 e ^ { 0.01331 t }$; population in $2000 : 8.1$ million
C) $P ( t ) = 2.1 e ^ { 0.10649 t }$; population in 2000: $2.3$ million
D) $P ( t ) = 2.1 e ^ { 0.10649 t }$; population in $2000 : 2.1$ million

Accepted Answer

The given function is $P(t) = 8(1.0134)^t$. To convert this to a function with base $e$, we use the formula $P(t) = P_0e^{rt}$, where $e$ is the base of the natural logarithm. We need to find $r$ such that $e^r = 1.0134$. Taking the natural logarithm of both sides, we get $r = \ln(1.0134)$, which approximately equals $0.01331$ when rounded to five decimal places. Therefore, the equivalent function is $P(t) = 8e^{0.01331t}$. For the year 2000, or $t = 0$, $P(0) = 8(1.0134)^0 = 8$, so the population in 2000 was 8 million.

Question 4

Solve the problem.
-If the half-life of an element is $67 \mathrm { yr }$ and the initial quantity is $3 \mathrm {~kg}$, write a function of the form $Q ( t ) = Q _ { 0 } e ^ { - k t }$ to model the quantity of the element left after $t$ years. Round $k$ to 4 decimal places.
A) $Q ( t ) = 67 e ^ { - 0.0103 t }$
B) $Q ( t ) = 67 e ^ { - 0.0034 t }$
C) $Q ( t ) = 3 e ^ { - 0.0034 t }$
D) $Q ( t ) = 3 e ^ { - 0.0103 t }$

Accepted Answer

The half-life formula is $Q(t) = Q_0 e^{-kt}$, where $Q_0$ is the initial quantity, $k$ is the decay constant, and $t$ is time. For a half-life of 67 years, $Q(t) = \frac{Q_0}{2}$ when $t = 67$. Substituting these values into the formula gives $0.5 = e^{-67k}$. Solving for $k$ gives $k \approx 0.0103$. Therefore, the correct model with $Q_0 = 3 \, \text{kg}$ is $Q(t) = 3e^{-0.0103t}$.

Question 5

Solve the problem.
-After a new product is launched the cumulative sales $S ( t )$ (in $\$ 1000 ) t$ weeks after launch is given by:
$$S ( t ) = \frac { 64 } { 1 + 8 e ^ { - 0.47 t } }$$
What is the limiting value in sales?
A) $\$ 8,000$
B) $\$ 64,000$
C) $\$ 6,000$
D) $\$ 47,000$

Accepted Answer

The limiting value in sales is found by taking the limit of $S(t)$ as $t$ approaches infinity. As $t$ goes to infinity, $e^{-0.47t}$ approaches 0, making the denominator $1 + 8(0) = 1$. Thus, the limiting value is $S(t) = \frac{64}{1} = 64$ in $1000, which is $\$64,000$.

Question 6

Solve the problem.
-The population of a country on January 1, 2000, is $20.7$ million and on January 1, 2010, it has risen to $22.6$ million. Write a function of the form $P ( t ) = P _ { 0 } e ^ { r t }$ to model the population $P ( t )$ (in millions) $t$ years after January 1, 2000. Then use the model to predict the population of the country on January $1,2,017$. round to the nearest hundred thousand.
A) $P = 20.7 e ^ { 0.00878 t } ; 24$ million
B) $P = 22.6 e ^ { 0.00878 t } ; 130$ million
C) $P = 22.6 e ^ { 0.00878 t } ; 26.2$ million
D) $P = 20.7 e ^ { 0.00878 t } ; 119$ million

Accepted Answer

The function $P(t) = 20.7e^{rt}$ models the population $t$ years after 2000 with $P_0 = 20.7$ million. To find $r$, use $P(10) = 22.6$ million: $22.6 = 20.7e^{10r}$. Solving for $r$ gives $r \approx 0.00878$. To predict the population in 2017 ($t = 17$), substitute $t = 17$ into the model to get $P(17) \approx 24$ million.

Question 7

Solve the problem.
-Eight million E. coli bacteria are present in a laboratory culture. An antibacterial agent is introduced and the population of bacteria $P ( t )$ decreases by half every $5 \mathrm { hr }$. The population can be represented by $P ( t ) = 8,000,000 \left( \frac { 1 } { 2 } \right) ^ { t / 5 }$. Convert this to an exponential function using base $e$.
A) $P ( t ) = 8,000,000 e ^ { - 0.13863 t }$
B) $P ( t ) = 4,000,000 e ^ { - 0.13863 t }$
C) $P ( t ) = 8,000,000 e ^ { - 0.1 t }$
D) $P ( t ) = 4,000,000 e ^ { - 0.1 t }$

Accepted Answer

The given function $P(t) = 8,000,000 \left( \frac{1}{2} \right)^{t/5}$ can be converted to an exponential function with base $e$ by using the identity $a^x = e^{x \ln(a)}$. Here, $a = \frac{1}{2}$ and $x = \frac{t}{5}$, so the function becomes $P(t) = 8,000,000 e^{\frac{t}{5} \ln(\frac{1}{2})}$. Since $\ln(\frac{1}{2})$ is approximately $-0.693147$, the exponent becomes $\frac{t}{5} \times -0.693147 = -0.13863t$, resulting in $P(t) = 8,000,000 e^{-0.13863t}$.

Question 8

Solve the problem.
-Suppose that $P$ dollars in principal is invested in an account earning $5.3 \%$ interest compounded continuously. At the end of $3 \mathrm { yr }$, the amount in the account has earned $\$ 1,378.70$ in interest. Find the original principal. Round to the nearest dollar. (Hint: Use the model $A = P e ^ { r t }$ and substitute $P + 1,378.70$ for $A$.)
A) $\$ 9,379$
B) $\$ 2,555$
C) $\$ 1,176$
D) $\$ 8,000$

Accepted Answer

The answer of Solve the problem.
-Suppose that $P$ dollars in...

Question 9

Solve the problem.
-If the function $Q ( t ) = 4 e ^ { - 0.00938 t }$ models the quantity (in $\mathrm { kg }$ ) of an element in a storage unit after $t$ years, how long will it be before the quantity is less than $1.5 \mathrm {~kg}$ ? Round to the nearest year.
A) $105 \mathrm { yr }$
B) $31 \mathrm { yr }$
C) $45 \mathrm { yr }$
D) $4 \mathrm { yr }$

Accepted Answer

To find when the quantity is less than 1.5 kg, set $Q(t) < 1.5$ and solve for $t$:$$4e^{-0.00938t} < 1.5$$Divide both sides by 4:$$e^{-0.00938t} < \frac{1.5}{4}$$$$e^{-0.00938t} < 0.375$$Take the natural logarithm of both sides to solve for $t$:$$-0.00938t < \ln(0.375)$$$$t > \frac{\ln(0.375)}{-0.00938}$$$$t > 104.72$$Rounding to the nearest year, it will be 105 years before the quantity is less than 1.5 kg.

Question 10

Solve the problem.
-The sales of a book tend to increase over the short-term as word-of-mouth makes the book "catch on." The number of books sold N(t) for a new novel t weeks after release at a certain book store is given In the table for the first 6 weeks. 
$\begin{array}{c|c}
\text { Weeks } t & \begin{array}{c}
\text { Number } \
\text { Sold } N(t)
\end{array} \
\hline 1 & 20 \
2 & 28 \
3 & 33 \
4 & 36 \
5 & 38 \
6 & 41
\end{array}$

Book Sales vs. Weeks After Release

a. Find a model of the form $y = a + b \ln t$.
b. Use the model to predict the sales in week 11. Round to the nearest whole unit.
c. Is it reasonable to assume that the logarithmic trend will continue? Why or why not?
A) a. $y = 3.9 + 19 \ln t$
b. 49 books
c. Yes. The number of books sold will continue to increase although the sales will approach a limit.
B) a. $y = 20.1 + 11.5 \ln t$
b. 48 books
c. Yes. The number of books sold will continue to increase although the sales will approach a limit.
C) a. $y = 3.9 + 19 \ln t$
b. 49 books
c. No. The trend cannot continue indefinitely. At some point, book sales will begin to decrease as most reading enthusiasts have read the book. 
D) a. $y = 20.1 + 11.5 \ln t$
b. 48 books
c. No. The trend cannot continue indefinitely. At some point, book sales will begin to decrease as most reading enthusiasts have read the book.

Accepted Answer

The answer of Solve the problem.
-The sales of a book...

Question 11

Graph the points and from visual inspection, select the model that would best fit the data. Choose from $\begin{array} { l l } 
y = m x + b \text { (linear) } & y = a b ^ { x } \text { (exponential) } \
y = a + b \ln x \text { (logarithmic) } & y = \frac { c } { 1 + a e ^ { - b x } } \text { (logistic) }
\end{array}$ Then use a graphing utility to find a function that fits the data. (Hint: For a logistic model, go to STAT, CALC,
Logistic.)
-$\begin{array}{c|c}
x & y \
\hline 10 & 41.2 \
20 & 47.5 \
30 & 51.2 \
40 & 53.8 \
50 & 55.9 \
60 & 57.5
\end{array}$

A) $y = 0.31 x + 40.3$
B) $y = \frac { 138 } { 1 + 0.23 e ^ { - 0.09 x } }$
C) $y = 40.8 ( 1.01 ) ^ { x }$
D) $y = 20.2 + 9.11 \ln x$

Accepted Answer

The answer of Graph the points and from visual inspection,...

Question 12

Solve the problem.
-After a new product is launched the cumulative sales $S ( t )$ (in $\$ 1000$ ) $t$ weeks after launch is given by:
$$S ( t ) = \frac { 74 } { 1 + 14 e ^ { - 0.4 t } }$$
Determine the cumulative amount in sales 4 weeks after launch. Round to the nearest thousand.
A) $\$ 19,000$
B) $\$ 1,000$
C) $\$ 55,000$
D) $\$ 10,000$

Accepted Answer

The answer of Solve the problem.
-After a new product is...

Question 13

Solve the problem.
-After a new product is launched the cumulative sales $S ( t )$ (in $\$ 1000 ) t$ weeks after launch is given by
$$S ( t ) = \frac { 54 } { 1 + 11 e ^ { - 0.32 t } }$$
Determine the amount of time for the cumulative sales to reach $\$ 48,000$. Round to the nearest week.
A) 14 weeks
B) 1 week
C) 27 weeks
D) 54 weeks

Accepted Answer

To find the time $ t $ when sales reach \$48,000, set $ S(t) = 48 $ (since sales are in thousands) and solve for $ t $:  $$ 48 = \frac{54}{1 + 11e^{-0.32t}} $$  Solving this equation gives $ t \approx 14 $ weeks.

Question 14

Solve the problem.
-The monthly costs for a small company to do business has been increasing over time in part due to inflation. The table gives the monthly cost y (in $) for the month of January for selected years. The Variable t represents the number of years since 2008.
 $\begin{array}{c|c}
\begin{array}{c}
\text { Year } \
(t=\mathbf{0} \text { is 2008) }
\end{array} & \begin{array}{c}
\text { Monthly } \
\text { Costs (\$) } y
\end{array} \
\hline 0 & 14,000 \
1 & 14,200 \
2 & 14,400 \
3 & 14,600
\end{array}$

Monthly Cost (Jan.) for Selected Years

Year $( \mathrm { t } = 0$ represents $2008 )$

a. Use a graphing utility to find a model of the form $y = a b ^ { t }$.
b. Write the function from part (a) as an exponential function with base $e$.
c. Use the model to predict the monthly cost for January in the year 2,017 if this trend continues. Round to the nearest hundred dollars.

A) a. $y = 14,001 ( 0.0140 ) ^ { t }$
b. $y = 10 e ^ { 0.0140 t }$
c. $\$ 91,900$

B) a. $y = 14,001 ( 1.014 ) ^ { t }$
b. $y = 14,001 e ^ { 0.0140 t }$
c. $\$ 15,900$

C) a. $y = 14,001 ( 1.014 ) ^ { t }$
b. $y = 10 e ^ { 0.0140 t }$
c. $\$ 15,900$

D) a. $y = 14,001 ( 1.014 ) ^ { t }$
b. $y = 10 e ^ { 1.014 t }$
c. $\$ 91,900$

Accepted Answer

The answer of Solve the problem.
-The monthly costs for a...

Quiz 4: Exponential and Logarithmic Functions

Solve the problem. -If the half-life of an element is $67 \mathrm { yr }$ and the initial quantity is $3 \mathrm {~kg}$ , write a function of the form $Q ( t ) = Q _ { 0 } e ^ { - k t }$ to model the quantity of the element left after $t$ years. Round $k$ to 4 decimal places.

Solve the problem. -After a new product is launched the cumulative sales $S ( t )$ (in $\$ 1000 ) t$ weeks after launch is given by: $S ( t ) = \frac { 64 } { 1 + 8 e ^ { - 0.47 t } }$ What is the limiting value in sales?

Solve the problem. -If the function $Q ( t ) = 4 e ^ { - 0.00938 t }$ models the quantity (in $\mathrm { kg }$ ) of an element in a storage unit after $t$ years, how long will it be before the quantity is less than $1.5 \mathrm {~kg}$ ? Round to the nearest year.

Solve the problem. -After a new product is launched the cumulative sales $S ( t )$ (in $\$ 1000$ ) $t$ weeks after launch is given by: $S ( t ) = \frac { 74 } { 1 + 14 e ^ { - 0.4 t } }$ Determine the cumulative amount in sales 4 weeks after launch. Round to the nearest thousand.

Solve the problem. -After a new product is launched the cumulative sales $S ( t )$ (in $\$ 1000 ) t$ weeks after launch is given by $S ( t ) = \frac { 54 } { 1 + 11 e ^ { - 0.32 t } }$ Determine the amount of time for the cumulative sales to reach $\$ 48,000$ . Round to the nearest week.

Quiz 4: Exponential and Logarithmic Functions

Solve the problem. -After a new product is launched the cumulative sales S(t) S ( t ) S(t) (in $1000) t\$ 1000 ) t$1000) t weeks after launch is given by: S(t) =641+8e−0.47tS ( t ) = \frac { 64 } { 1 + 8 e ^ { - 0.47 t } }S(t) =1+8e−0.47t64​ What is the limiting value in sales?

Solve the problem. -After a new product is launched the cumulative sales $S ( t )$ (in $\$ 1000 ) t$ weeks after launch is given by: $S ( t ) = \frac { 64 } { 1 + 8 e ^ { - 0.47 t } }$ What is the limiting value in sales?