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The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

# Exponential Growth

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

## Objectives

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
• $N_i$ = 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
• $A$ = 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)

## Formulas

(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)}} #### For Continuous Compound Interest: (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} ## Solved Examples 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\: yearsN_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 $N_f =$ $$2376 k = 0.061387865 (it is a constant) N_i =$$269t = ? 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 ## Calculators for Exponential Growth Given: growth rate To Find: doubling time in Given: doubling time To Find: growth rate Required: initial amount, growth rate Optional: time To Find: other details in in "whatever unit" Required: initial amount, growth rate Optional: future amount To Find: other details in Given: initial amount, time, doubling time To Find: other details in "whatever unit" in "the same whatever unit" Given: initial amount, time, future amount To Find: other details in "whatever unit" #### References Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com Bittinger, M. L., Beecher, J. A., Ellenbogen, D. J., & Penna, J. A. (2017). Algebra and Trigonometry: Graphs and Models ($6^{th}$ed.). Boston: Pearson. Sullivan, M., & Sullivan, M. (2017). Algebra & Trigonometry ($7^{th}\$ ed.). Boston: Pearson.

Authority (NZQA), (n.d.). Mathematics and Statistics subject resources. www.nzqa.govt.nz. Retrieved December 14, 2020, from https://www.nzqa.govt.nz/ncea/subjects/mathematics/levels/

CMAT Question Papers CMAT Previous Year Question Bank - Careerindia. (n.d.). Retrieved May 30, 2020, from https://www.careerindia.com/entrance-exam/cmat-question-papers-e23.html