Here, we need to use calculator to express the answer to the nearest hundredth (2 decimal places).

a) $A=250(1.05)^{10}=407.22$

b) $P=9000left(dfrac{1}{2}right)^8=35.16$

c) $N_0=dfrac{500}{(1.25)^{1.25}}=378.30$

d) $P=dfrac{625}{0.71^9}=13631.85$

$bold{Concept:}$ For an exponential function $y=ab^x$ where $xgeq 0$

$a=$ the initial amount or number

growth if $b>1$, growth rate = $b-1$

decay if $0<b<1$, decay rate = $1-b$

begin{table}[]
defarraystretch{1.8}%
begin{tabular}{|l|l|l|l|}
hline
Function & begin{tabular}[c]{@{}l@{}}Exponential Growth\ or Decay?end{tabular} & Initial Value & begin{tabular}[c]{@{}l@{}}Growth or \ Decay Rateend{tabular} \ hline
a) $V=20(1.02)^t$ & $1.02>1implies$ growth & 20 & $1.02-1=0.02=2%$ \ hline
b) $P=1(0.8)^n$ & $0.81implies$ growth & 0.5 & $3-1=2=200%$ \ hline
d) $Q=600left(frac{5}{8}right)^w$ & $frac{5}{8}<1implies$ decay & 600 & $frac{5}{8}-1=-0.375=-37.5%$ \ hline
end{tabular}
end{table}

begin{table}[]
defarraystretch{1.4}%
begin{tabular}{|l|l|l|l|}
hline
Function & begin{tabular}[c]{@{}l@{}}Exponential Growth\ or Decay?end{tabular} & Initial Value & begin{tabular}[c]{@{}l@{}}Growth or \ Decay Rateend{tabular} \ hline
a) $V=20(1.02)^t$ & growth & 20 & $2%$ \ hline
b) $P=1(0.8)^n$ & decay & 1 & $-20%$ \ hline
c) $A=0.5(3^x)$ & growth & 0.5 & $200%$ \ hline
d) $Q=600left(frac{5}{8}right)^w$ & decay & 600 & $-37.5%$ \ hline
end{tabular}
end{table}

$bold{Concept:}$ For an exponential function $y=ab^x$ where $xgeq 0$

$a=$ the initial amount or number

growth if $|b|>1$, growth rate = $b-1$

decay if $0<b<1$, decay rate = $b-1$

$bold{Solution:}$

a) The initial population is the value of $P$ when $n=0$ which is $1250$ which is also the value of $a$ for the general exponential function $y=ab^x$

b) The growth rate is $1.03-1=0.03=3%$ which is the value of $b-1$ in the general exponential function.

c) In the year 2007, $n=2007-1996=11$

$P(11)=1250(1.03)^{11}=1730$

Therefore, the population is 1730 in the year 2007.

d) We shall find $n$ such that $P(n)=2000$

$2000=1250(1.03)^n$

$1.03^n=dfrac{2000}{1250}$

$1.03^n=1.6$

We shall solve this graphically.

Therefore, the population would reach 2000 in about 16 years after 1996 which is

$$
1996+16=2012
$$

a) 1250 ; the value of $P$ when $n=0$

b) $3%$ ; the value of $b-1$ in the general exponential function

c) 1730

d) 2012

$bold{Concept:}$ For a general exponential function $y=ab^x$ where $xgeq 0$

$a=$ the initial amount or number

growth or decay rate = $b-1$

growth if $b>1$

decay if $0<b<1$

$bold{Solution:}$ We are given the value as a function of time $V(m)=1500(0.95)^m$

a) The initial value occurs when $m=0$ so the initial value is $V(0)=1500(0.95)^0=1500$

We can also see that $V(m)$ is a general exponential function with $a=1500$ which corresponds to the initial value.

b) The rate of depreciation is the decay rate. In this case, $b=0.95$

decay rate= $b-1=0.95-1=-0.05=-5%$

c) The value of the computer after 2 years (24 months) corresponds to the value of $V$ when $m=24$

$V(2)=1500(0.95)^{24}=$437.98$

d) Here, we shall solve for $m$ such that $V(m)=900$

$900=1500(0.95)^m$

$0.95^m=dfrac{900}{1500}$

We shall solve this graphically and we get $m=10$ months

a) $$1500$, it is the value of $a$ in the general exponential function

b) $5%$, it is the value of $b-1$

c) $$437.98$

d) $10^{text{th}}$ month after purchase

$bold{Concept:}$ For a general exponential function $y=ab^x$ where $xgeq 0$

$a=$ the initial amount or number

growth or decay rate = $b-1$

growth if $b>1$

decay if $0<b<1$

$bold{Solution;:}$ The initial amount invested is $$1000$ with a growth rate of $6%=0.06$. We can model this using the general exponential function $y=ab^x$ where $a=1000$ and $b=1+0.06=1.06$

a) growth rate = $6%$

b) $$1,000$

c) $15$ periods

d) If $V$ is the future amount and the $n$ is the number of years, the equation would be

$V=1000(1.06)^n$

The amount after $15$ years is the value of $V$ at $n=15$

$V(15)=1000(1.06)^{15}=$2396.56$

a) $6%$

b) $$1,000$

c) $15$

d) $V=1000(1.06)^n$ , $$2396.56$

a) The exponent is $dfrac{1}{10}$ because the population doubles every 10 hours.

b) The base is 2 because the bacteria $bold{doubles}$ (100$%$ growth rate) every 10 hours.

c) The multiplier is 500 because it corresponds to the initial population at $t=0$

d) At Midnight, 12 hours have passed from noon, so

$P(12)=500left(2^{frac{12}{10}}right)=bold{1149;text{bacteria}}$

e) At noon the next day, $t=24$ (1 day = 24 hours)

$P(24)=500left(2^{frac{24}{10}}right)=bold{2639}$ $bold{text{bacteria}}$

f) We need to find $t$ such that $P(t)>2000$

$2000=500left(2^{frac{t}{10}}right)$

By graphical approach, we get $t=20$ hours which correspond to $bf {8:00; am }$ the next day.

a) The population doubles in $bold{10}$ hours.

b) The population $bold{doubles}$ every 10 hours.

c) 500 corresponds to the initial population

d) 1149 bacteria

e) 2639 bacteria

f) 8:00 am the next day

$bold{Concept:}$ For a general exponential function $y=ab^x$ where $xgeq 0$

$a=$ the initial amount or number

growth or decay rate = $b-1$

$bold{exponential; growth}$ if $b>1$

$bold{exponential; decay}$ if $0<b<1$

$bold{Solution:}$ The given functions are of the form $y=ab^x$ so we shall examine the value of $b$ to determine whether it is an exponential decay.

a) $b=3>1implies$ exponential growth

b) $b=1.2>1implies$ exponential growth

c) $b=0.8<1implies$ exponential decay

d) $b=0.9<1implies$ exponential decay

Therefore, the functions that describe exponential decay are options (c) and (d)

The population function is given as

$P(n)=12(1.025)^n$

where $P(n)$ is the population in thousands and $n$ is the number of years from now.

a) The population of the town in 10 years corresponds to the value of $P$ when $n=10$

$P(10)=12(1.025)^{10}=15.361implies 15361$ inhabitants

b) The population doubles when $P(n)=24$

$24=12(1.025)^n$

From the intersection of the graph $y=24$ and $y=12(1.025)^n$, we get $n=28.071$

So it will double in the 29th year.

c) We can use graphical approach to find the number of years ago that the population is $8000$. From the intersection point between $y=8$ and $y=12(1.025)^n$, we see that it happened around $16$ years ago.

d) The domain of the function is all real numbers, that is, no real number can make the function undefined.

domain = ${ ninbold{R}}$

The function is increasing and has an asymptote at $y=0$. So the range is

range = ${ Pinbold{R};|;P>0}$

a) 15361

b) 29th year

c) 16 years ago

d) ${ ninbold{R}}; ;;{ Pinbold{R};|;Pgeq 0}$

We are given with the internal temperature as a function of time

$T(t)=63(0.5)^{frac{t}{10}}+19$

a) The temperature when she began to record its temperature is the value of $T$ at $t=0$

$T(0)=63(0.5)^{frac{0}{10}}+19=63(1)+19=82^circ$ C

b) The temperature after 20 minutes is the value of $T$ when $t=20$

$T(20)=63(0.5)^{frac{20}{10}}+19=34.75approx 35^circ$ C

c) The time it takes to reach $T=30^circ$ can be obtained graphically.

From the intersection point of $y=30$ and $y=63(0.5)^{frac{x}{10}}+19$, it would take 25 minutes to reach $30^circ$

a) $82^circ$C

b) $35^circ$C

c) after 25 minutes

$bold{Concept:}$ The general exponential model is

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$

$bold{Solution:}$

a) Let $C$ be the color after washing $n$ times. Since $C$ is in percent, the initial value is 100

“lose 1$%$ of their color” $implies r=-1%=-0.01$

$C=100(1+(-0.01))^n$

$C=100(0.99)^n$

b) Let $P$ be the population $t$ years after 1990 and $P_0=2500$ be the initial population.

“grew at a rate of 0.5$%$ each year” $implies r=0.05%=0.005$

$P=2500(1+0.005)^{t}$

$P=2500(1.005)^t$

c) Let $P$ be the bacterial population after $t$ days and $P_0$ is the initial population.

“1 day to divide into two” $implies$ “doubles every day” $implies r=100%=1$

$P=P_0(1+1)^{t}$

$P=P_0(2)^t$

a) $C=100(0.99)^n$

b) $P=2500(1.005)^t$

c) $P=P_0(2)^t$

a) Since the population doubles per hour, the growth rate is $100%$ per hour.

b) The population after $t$ hours is an exponential function with base of 2 and a coefficient of 80 (the initial population)

$P=80(2)^t$

c) The population after 6 hours is the value of $P$ at $t=6$

$P(6)=80(2)^{6}=5120$

d) To get the population after 90 minutes, we must convert to hours

$90 ;text{min} times dfrac{1;text{hour}}{60;text{min}}=1.5$ hours

$P(1.5)=80(2)^{1.5}=226$

e) We can find the number of hours for it to reach 1 million by graphing.

From the intersection point, it would take $13.61$ hours for it to reach 1 million.

f) The domain of exponential functions are all real numbers, however in this situation, the domain is absolute time which cannot be negative. Thus the domain is the set of non-negative real numbers.

domain = ${ tinbold{R};|;tgeq 0}$

Since it is an exponential growth, the population cannot be less than the initial population

range = ${ Pinbold{R};|;pgeq 80}$

a) 100$%$

b) $P=80(2^t)$

c) $5120$

d) 226

e) 13.6 hours

f) ${ tinbold{R};|;tgeq 0}$ ; ${ Pinbold{R};|;pgeq 80}$

$bold{Concept:}$ The general exponential model is

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$

$bold{Solution:}$

a) Let $n$ be the number of years after 1990. We are given that the initial value $V_0$ is 5 and the growth rate $r=6%=0.06$. Using the exponential model, the value of the card after $t$ years $V(t)$ is

$V=V_0(1+r)^t$

$V=5(1+0.06)^t$

$V=5(1.06)^t$

b) The increase in value from year 3 to year 4 is $V(4)-V(3)$

$V(4)-V(3)=5(1.06)^4-5(1.06)^3=$0.36$

c) The increase in value in the 20th year is $V(20)-V(19)$

$$
V(20)-V(19)=5(1.06)^{20}-5(1.06)^{19}=$0.91
$$

a) $V=5(1.06)^t$

b) $$0.36$

c) $$0.91$

$bold{Concept:}$ The general exponential model is

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$ per period

$bold{Solution:}$

a) Let $I$ be the intensity at any depth $d$ in meters

“falls by 9% per metre” $implies r=-0.09$

“intensity is 100$%$ at the surface” $implies I_0=100$

Therefore,

$I=I_0(1+r)^d$

$I=100(1-0.09)^{d}$

$I=100(0.91^d)$

b) We shall determine the value of $I$ when $d=7.5$

$$
I=100(0.91^7.5)=49.3%
$$

a) $I=100(0.91^d)$

b) $49.3%$

$bold{Concept:}$ The general exponential model is

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population/quantity after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$ per period

$bold{Solution:}$

a) Let $P$ be the population of germs in $%$ and $a$ be the number of applications.

“kill 99$%$ of all germs each application” $implies r=-0.99$

Initially, $100%$ is present so $P_0=100$

$P=P_0(1+r)^a$

$P=100(1-0.99)^a$

$P=100(0.01^a)$

b) Let $N$ be the number of germs present so that $N_0=10^{10}$

$N=10^{10}(0.01^a)$

All germs are eliminated when $N<1$

$10^{10}(0.01^a)<1$

$0.01^a<10^{-10}$

Notice that $0.01=10^{-2}$

$10^{-2a}<10^{-10}$

$-2a10$

$a>5$

Thus, it must applied more than 5 times which is 6 times.

a) $P=100(0.01)^a$

b) 6 times

$bold{Concept:}$ The general exponential model is

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$ per period

$bold{Solution:}$

Let $P$ be the population of the town at any year, $P_0$ be the population in 1990, and $n$ be the number of years after $1990$.

We have the following data:

$P_0=8400$

$P(15)=12;500$

Use the general exponential model

$P(15)=P_0(1+r)^{15}$

$12;500=8400(1+r)^{15}$

$(1+r)^{15}=dfrac{12;500}{8400}$

$1+r=left(dfrac{12;500}{8400}right)^{frac{1}{15}}$

$1+r=1.027$

$r=0.027=2.7%$

The growth rate is approximately $2.7%$

approximately $2.7%$

$bold{Concept:}$ The general exponential model is

$A(n)=A_0 (1+r)^{n}$

$A(n)$ is the amount/population after $n$ periods

$A_0$ is the initial amount

$r$ is the growth rate $(text{if} ;r>0)$ or decay rate $(text{if}; r<0)$ per period

$bold{Solution:}$

a) Let $P$ the population of the yeast cells , $P_0$ be the initial population, and $t$ be the number of hours elapsed.

“cell grows by 75$%$” $implies r=0.75$

“every 3 h” $implies$ when $t=3$, $n=1$ (one period) $implies$ $n = dfrac{t}{3}$

Using the general exponential model

$P=P_0(1+r)^n$

$P=200(1+0.75)^{frac{t}{3}}$

$P=200(1.75)^{frac{t}{3}}$

b) $P$ is the population after $t$ hours

$200$ is the initial population at 9 a.m.

$0.75$ is the growth rate

$frac{t}{3}$ specifies that the cell grow every 3 h

a) $P=200left(1.75^{frac{t}{3}}right)$

b) $dfrac{t}{3}$

a) Let $x$ be the number of years after 2002 and $y$ be the number of babies named Nevaeh each year.

Examine the given pattern

$x=0$ , $y=1$

$x=2$ , $y=18$

$x=3$ , $y=70$

Use the general exponential function and find $b$

$y=ab^x$

$18=1(b)^2implies b=sqrt{18}= 4.243$

$70=1(b)^3implies b=sqrt[3]{70}=4.121$

We see that $b$ is relatively close in value for both points.

Thus, it could be modeled as exponential growth.

b) We could use $y=4.24^x$ or $y=4.12^x$ to model this

a) Exponential growth could model the situation.

b) $y=4.24^x$ or $y=4.12^x$ could fit the data points

c) There are also many factors that could change the exponential trend. For example, when so many girls have this name, parents are not likely to name their daughter as such leading to a decline in the growth rate.

a) The graph can be sketched as follows.

The graph sharply decreases at small values of $T$ and the rate slows down for large values of $T$ (asymptotic) which is a characteristic of exponential decay.

The function itself is NOT exponential since the independent variable is not at the exponent, however the behavior could be modeled by exponential decay.

b) Find the value of $R$ when $T=24$

$$
R(24)=dfrac{100}{1+1.08(24)^{0.21}}=32.2%
$$

a) The given function is not exponential, however its behavior could be modeled by exponential decay.

b) about $32.2%$