Exponential Growth

Pre-requisites: Exponents and Logarithms

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Use at least two methods whenever applicable.
Show all work

(1.) In 2012, the population of Samdom For Peace city was 6.03 million.
The exponential growth rate was 1.84% per year.
(a.) Determine the exponential growth function.
(b.) Estimate the population of the city in 2018. Round to the nearest tenth as needed.
(c.) When will the population of the city be 10 million? Round to the nearest tenth as needed.
(d.) Calculate the doubling time. Round to the nearest tenth as needed.

$ N_f = N_ie^{kt} \\[3ex] N_i = 6.03 \\[3ex] k = 1.84\% = 0.0184 \\[3ex] N_f = 6.03e^{0.0184t} \\[5ex] In\: 2018, \\[2ex] t = 2018 - 2012 = 6 \\[3ex] N_f = 6.03 * e^{0.0184 * 6} \\[3ex] N_f = 6.03 * e^{0.1104} \\[3ex] N_f = 6.03 * 1.116724671 \\[3ex] N_f = 6.733849766 \\[3ex] N_f = 6.7\:\: million \\[5ex] When\: N_f = 10, \\[2ex] t = \dfrac{\ln{\left(\dfrac{N_f}{N_i}\right)}}{k} \\[5ex] t = \dfrac{\ln{\left(\dfrac{10}{6.03}\right)}}{0.0184} \\[5ex] t = \dfrac{\ln{(1.658374793)}}{0.0184} \\[5ex] t = \dfrac{0.505838082}{0.0184} \\[5ex] t = 27.49120012 \\[2ex] t = 27.5\:\: years \\[5ex] T_{double} = \dfrac{\ln 2}{k} \\[5ex] T_{double} = \dfrac{0.693147181}{0.0184} \\[5ex] T_{double} = 37.67104242 \\[2ex] T_{double} = 37.7\:\: years $

(2.) A toy tractor sold for $$269$ in $1980$ and was sold again in $1990$ for $$497$.
Assume that the growth in the value of the collector's item was exponential.
Calculate the amount of time after which the value of the toy tractor will be $$2376$?
Round to two decimal places as needed.
Then, round to the nearest integer.
What year will that be? Interpret it.

In this case, we shall let $1980$ be the initial year.

From $1980$ to $1990$, the time, $t = 1990 - 1980 = 10\: years$

$N_i =$ $$269$

$N_f =$ $$497$

Let us calculate the growth constant.

$ k = \dfrac{\ln{\left(\dfrac{N_f}{N_i}\right)}}{t} \\[5ex] k = \dfrac{\ln{\left(\dfrac{497}{269}\right)}}{10} \\[5ex] k = \dfrac{\ln 1.847583643}{10} \\[5ex] k = \dfrac{0.613878646}{10} \\[5ex] k = 0.061387865 $
When $N_f =$ $$2376$

$k = 0.061387865$ (it is a constant)

$N_i =$ $$269$

$t = ?$

$ t = \dfrac{\ln{\left(\dfrac{N_f}{N_i}\right)}}{k} \\[5ex] t = \dfrac{\ln{\left(\dfrac{2376}{269}\right)}}{0.061387865} \\[5ex] t = \dfrac{\ln{8.832713755}}{0.061387865} \\[5ex] t = \dfrac{2.178462301}{0.061387865} \\[5ex] t = 35.48685560 \\[2ex] t = 35.49\: years $
$t = 35\: years$

The year will be $1980 + 35 = 2015$

This means that in the year, $2015$; the tractor is expected to sell for two thousand, three hundred and seventy six dollars.

(3.) ACT Observation of a certain bacteria colony has shown that its population of cells doubles every $3$ hours.
Given that the initial population of cells in this colony is about $8$ million, which of the following values, in millions, would be closest to the number of cells in the bacteria colony after $15$ hours?

We shall do this in three ways.
Choose any way you prefer.
ACT is a timed test...$60$ minutes to solve $60$ questions.
Ideally, each question should be solved in approximately a minute.
However, some questions take less than a minute to solve.
This means that the minutes saved from solving those questions should be used to solve questions that takes more than a minute.
The first method is recommended for the ACT
The first method is also recommended if you do not know the formula or if you were not given the formula.

First Method: Quantitative Reasoning

The population doubles every $3$ hours.
Initial population is $8$ million.
After the first $3$ hours, the population will be $8 * 2 = 16$ million
After the second $3$ hours ($6$ hours), the population will be $16 * 2 = 32$ million
After the third $3$ hours ($9$ hours), the population will be $32 * 2 = 64$ million
After the fourth $3$ hours ($12$ hours), the population will be $64 * 2 = 128$ million
After the fifth $3$ hours ($15$ hours), the population will be $128 * 2 = 256$ million

Second Method: Formula

$ T_{double} = 3 \:\:hours \\[3ex] N_i = 8 \:\:million \\[3ex] t = 15 \:\:hours \\[3ex] N_f = ? \\[3ex] N_f = N_ie^{kt} \\[3ex] k = \dfrac{\ln 2}{T_{double}} \\[5ex] k = \dfrac{\ln 2}{3} \\[5ex] kt = \dfrac{\ln 2}{3} * 15 = \ln 2 * 5 = 5\ln 2 \\[5ex] \implies N_f = 8 * e^{5\ln 2} \\[3ex] N_f = 8 * e^{\ln 2^5} \\[3ex] N_f = 8 * e^{\ln 32} \\[3ex] e^{\ln 32} = 32 ...Law\:\: 7...Log \\[3ex] N_f = 8 * 32 \\[3ex] N_f = 256 \:\:million \\[3ex] $ Third Method: Formula

$ T_{double} = 3 \:\:hours \\[3ex] N_i = 8 \:\:million \\[3ex] t = 15 \:\:hours \\[3ex] N_f = ? $ $N_f = N_i * 2$^{$\dfrac{t}{T_{double}}$}

$ \dfrac{t}{T_{double}} = \dfrac{15}{3} = 5 \\[5ex] N_f = 8 * 2^5 \\[3ex] N_f = 8 * 32 \\[3ex] N_f = 256 \:\:million $

(4.) ACT The population of a particular town is modeled by the equation $P = 120,000(1.1)^t$ where $t$ is the number of years after January $1$, $2011$.
Based on the model, which of the following numbers is closest to the population of the town on January $1$, $2013$?

$ A.\:\: 132,000 \\[3ex] B.\:\: 145,000 \\[3ex] C.\:\: 160,000 \\[3ex] D.\:\: 264,000 \\[3ex] E.\:\: 396,000 $

$ P = 120,000(1.1)^t \\[3ex] t = 2013 - 2011 = 2 \\[3ex] P = 120,000(1.1)^2 \\[3ex] P = 145,200 \\[3ex] $ The closest is $145,000$

(5.) Assume the population of Love Thy Neighbor Country (wish it existed) has a population of four million, seven hundred and eighty three thousand people.
It has a land area of fourteen billion square yards.
It has an exponential growth rate of four and nine-tenths percent per year.
After how long will there be one person for every square yard of land?

$ Initial\:\:Population, N_i = 4783000 \\[3ex] Land\:\:Area = 14000000000\:\:square\:\:yards \\[3ex] Population\:\:Density = \dfrac{Initial\:\:Population}{Land\:\:Area} \\[5ex] Population\:\:Density = \dfrac{4783000}{14000000000} \\[5ex] Population\:\:Density = 0.000341642857\:\:people/square\:\:yard \\[3ex] But\:\:we\:\:need\:\:1\:person/square\:\:yard \\[3ex] This\:\:implies\:\:a\:\:Population\:\:Density\:\:of\:\:1\:\:person/square\:\:yard \\[3ex] This\:\:implies\:\:a\:\:population\:\:of\:\:14000000000 \\[3ex] \implies N_f = 14000000000 \\[3ex] When\:\:will\:\:that\:\:be\:\;means\:\:time? \\[3ex] time, t = ? \\[3ex] Growth\:\:rate, k = 4.9\% = \dfrac{4.9}{100} = 0.049 \\[5ex] t = \dfrac{\ln{\left(\dfrac{N_f}{N_i}\right)}}{k} \\[7ex] Numerator = \ln{\left(\dfrac{N_f}{N_i}\right)} = \ln{\left(\dfrac{14000000000}{4783000}\right)} \\[5ex] Numerator = \ln 2927.03324 \\[3ex] Numerator = 7.98174464 \\[3ex] Denominator = k = 0.049 \\[3ex] \rightarrow t = \dfrac{7.98174464}{0.049} \\[5ex] t = 162.892748\:\:years \\[3ex] t \approx 163\:\:years \\[3ex] $ It will take about a hundred and sixty three years for there to be one person per square yard of land.

(6.) The population of a herd of deer is represented by the function $A(t) = 200(1.31)^t$, where $t$ is given in years.
To the nearest whole number, what will the herd population be after 4 years?

$ A(t) = 200(1.31)^t \\[3ex] t = 4\;years \\[3ex] A(4) = 200(1.31)^4 \\[3ex] A(4) = 200(2.94499921) \\[3ex] A(4) = 588.999842 \\[3ex] $ The herd population will be approximately 588 after 4 years

(7.) The population of a city is modeled by the equation $P(t) = 354,584e^{0.2t}$ where $t$ is measured in years.
If the city continues to grow at this rate, how many years will it take for the population to reach one million?

$ P(t) = 354,584e^{0.2t} \\[3ex] P(t) = 1000000\;people \\[3ex] t = ? \\[3ex] 1000000 = 354584e^{0.2t} \\[3ex] 354584e^{0.2t} = 1000000 \\[3ex] e^{0.2t} = \dfrac{1000000}{354584} \\[5ex] e^{0.2t} = 2.820206213 \\[3ex] Introduce\;\;natural\;\;logarithm\;\;to\;\;both\;\;sides \\[3ex] \ln e^{0.2t} = \ln 2.820206213 \\[3ex] 0.2t = 1.036810008 \\[3ex] t = \dfrac{1.036810008}{0.2} \\[5ex] t = 5.184050038 \\[3ex] t \approx 5.18\;years \\[3ex] $ The population will reach a million people in about 5 years

(8.) (a.) During the exponential phase, E. coli bacteria in a culture increase in number at a rate proportional to the current population.
If the growth rate is 1.9% per minute and the current population is 172.0 million, what will the population be 7.2 minutes from now?

(b.) For a period of time, an island's population grows at a rate proportional to its population.
If the growth rate is 3.8% per year and the current population is 1543, what will the population be 5.2 years from now?

$ (a.) \\[3ex] N_f = N_ie^{kt} \\[3ex] N_i = 172\;million \\[3ex] k = 1.9\% = \dfrac{1.9}{100} = 0.019/minute \\[5ex] t = 7.2\;minutes \\[3ex] \implies \\[3ex] N_f = 172e^{0.019(7.2)} \\[3ex] = 172 \cdot e^{0.1368} \\[3ex] = 172(1.146598806) \\[3ex] = 197.21499 \\[3ex] \approx 197.2\;million \\[3ex] $ 7.2 minutes from now, the population will be approximately 197.2 million

$ (b.) \\[3ex] y = a(1 + r)^t \\[3ex] r = 3.8\% = \dfrac{3.8}{100} = 0.038/year \\[5ex] a = 1543 \\[3ex] t = 5.2\;years \\[3ex] \implies \\[3ex] y = 1543(1 + 0.038)^{5.2} \\[3ex] = 1543(1.038)^{5.2} \\[3ex] = 1543(1.214021109) \\[3ex] = 1873.235 \\[3ex] \approx 1874\;million \\[3ex] $ 5.2 years from now, the population will be approximately 1874 million

Student: Mr. C, why is it not approximately 1873?
Teacher: So, we rounded to the nearest integer because the current population was given as an integer
In rounding to the nearest integer, it is supposed to be approximately 1880
However, we are dealing with human beings in this case
If we were dealing with quantities such as money, then it is 1873
But when dealing with human beings, we have to put 1874 because a human being is a whole person.
Student: Are you saying that rounding depends on the quantity being rounded?
For human beings, irrespective of the integer after the decimal point, we should take it as 1 and add to the integer before the decimal point?
Teacher: That is correct.

(9.) Assume that in 1990, there were 184 alligators in Orange County, Florida.
The number of alligators increased by 72% per year after 1990.
How many alligators were there in Orange County, Florida in 2002?

$ N(t) = a(1 + r)^t \\[3ex] a = 184\;million \\[3ex] r = 72\% = \dfrac{72}{100} = 0.72/year \\[5ex] t = 2002 - 1990 = 12\;years \implies \\[3ex] N(12) = 184(1 + 0.72)^{12} \\[3ex] = 184(1.72)^{12} \\[3ex] = 184(670.4113572) \\[3ex] = 123,355.6897 \\[3ex] \approx 123,356 \\[3ex] $ At the rate of 75% per year, the population of alligators after 2002 was approximately 123,356.

Exponential Growth

Objectives

Symbols and Meanings

Formulas

For Growth Factor and Growth Rate

For Exponential Growth Rate and Doubling Time

For Continuous Compound Interest

Solved Examples

Calculators for Exponential Growth

References