Blessed be the GOD and FATHER of our LORD JESUS CHRIST, who has blessed us with all spiritual blessings in the heavenly places in CHRIST. Amen.

- Ephesians 1:3

The Joy of a Teacher is the Success of his Students.

- Samuel Dominic Chukwuemeka

I greet you this day,

__First:__ read the notes.

__Second:__ view the videos.

__Third:__ solve the **solved examples** and **word problems.**

__Fourth:__ check your solutions with my **thoroughly-explained** solutions.

__Fifth:__ check your answers with the calculators as applicable.

I wrote the codes for these calculators using JavaScript, a client-side scripting language.

Please use the latest Internet browsers. The calculators should work.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.

If you are my student, please do not contact me here. Contact me via the school's system.

Thank you for visiting.

**Samuel Dominic Chukwuemeka** (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

Students will:

(1.) Discuss exponential growth.

(2.) Discuss real-world applications of exponential growth.
(3.) Solve problems involving exponential growth.

## Symbols and Meanings

- To solve for a specified variable for each formula, please review Solved Examples - Literal Equations
- $N_i$ = initial amount (amount at time = $0$)
- $N_f$ = future amount (amount at time = $t$)
- $k$ = growth rate (growth constant)
- $t$ = time
- $T_{double}$ = doubling time
__For Continuous Compound Interest:__- You can verify all answers with the
**Mathematics of Finance Calculator** - $A$ = continuous compound amount or future value (in dollars, naira, any currency)
- $P$ = principal or present value (in dollars, naira, any currency)
- $r$ = annual interest rate (in percent)
- $t$ = time (in years)

(1.) $ N_f = N_ie^{kt} $

(2.) $N_f = N_i * 2$^{$\dfrac{t}{T_{double}}$}

(3.) $ t = \dfrac{\ln{\left(\dfrac{N_f}{N_i}\right)}}{k} $

(4.) $ k = \dfrac{\ln{\left(\dfrac{N_f}{N_i}\right)}}{t} $

(5.) $ T_{double} = \dfrac{\ln 2}{k} $

(6.) $ k = \dfrac{\ln 2}{T_{double}} $

(7.) $ T_{double} = \dfrac{t\ln2}{\ln{\left(\dfrac{N_f}{N_i}\right)}} $

(8.) $ A = Pe^{rt} $

(9.) $ P = \dfrac{A}{{e^{rt}}} $

(10.) $ t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} $

(11.) $ r = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{t} $

**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.

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

The first method is also recommended if you do not know the formula or if you were not given the formula.

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

$ 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] $

$ 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{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

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