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

# Exponential Growth

I greet you this day,
First: read the notes. Second: view the videos. Third: solve the questions/solved examples. 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

## 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 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. Solve each question. 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.

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