University:
Carleton College
Course:
MATH 111 | Introduction to Calculus
Academic year:

2024
Views:

420

Pages:

2
Author:

Dave D.

Algebra 2 Honors Problem Solving: Exponential Growth and Decay Name: Key Logs Notes 6) Warm Up Solve each of the following equations. a. 4^x = 1.7 x log 4 = log 1.7 x = log 1.7 / log 4 x ≈ 0.38 b. 4(x + 2)^5 = 16 (x + 2)^5 = 4 x + 2 = 1.3195 x = -0.68 Exponential Growth and Decay Models y = a(1 ± r)^t y: current amount a: initial amount r: rate as decimal (not %) t: time EXAMPLE 1 - Growth and Decay a. In 1996, there were 2573 computer viruses and other computer security incidents. During the next 7 years, the number of incidents increased by about 92% each year. A. Write an exponential growth model giving the number, n, of incidents t years after 1996. n = 2573(1 + 0.92)^t n = 2573(1.92)^t B. About how many incidents were there in 2003? n = 2573(1.92)^7 ≈ 247,485 viruses b. A new snowmobile costs $4200. The value of the snowmobile decreases by 10% each year. Find the value of the snowmobile after 3 years. y = 4200(1 - 0.10)^3 y = 4200(0.90)^3 y ≈ $3061.80 c. The value of a snowmobile has been decreasing by 7% each year since it was new. After 3 years, the value is 3000. Find the original cost of the snowmobile. 3000 = a(1 - 0.07)^3 3000 = a(0.93)^3 a ≈ $3729.69 d. The population of a city was 3,500,000 on April 1, 1980 and 5,000,000 on April 1, 1990. Assuming exponential growth, find the month and year that the population will be 6,500,000. 5000000 = 3500000(1 + r)^10 1.42857 = (1 + r)^10 1.0363 ≈ 1 + r r = 0.0363 6500000 = 3500000(1.0363)^t 1.857 = (1.0363)^t log 1.857 = t log 1.0363 t ≈ 17.355 months Month: August Year: 1997 Compound Interest Formula If an amount P (called the principal) is invested at an annual interest rate r compounded n times a year, then in t years the investment will grow to an amount A given by: A = P(1 + r/n)^(nt) Where r is expressed as a decimal. EXAMPLE 2 - Compounded Interest a. You deposit $4000 in an account that pays 2.92% annual interest. Find the balance after 1 year if the interest is compounded with the given frequency: A = 4000(1 + 0.0292/2)^(2*1) = $4117.65 (Semiannually) A = 4000(1 + 0.0292/4)^(4*1) = $4118.04 (Quarterly) A = 4000(1 + 0.0292/12)^(12*1) = $4118.38 (Monthly) A = 4000(1 + 0.0292/365)^(365*1) = $4118.52 (Daily) b. How many dollars must be invested at 16%, compounded quarterly, to yield $10,000 after 5 years? 10000 = P(1 + 0.16/4)^(4*5) 10000 = P(1.04)^20 P = $4563.87 c. How long will it take you to quadruple your money if you invest it at a rate of 3.5% compounded monthly? 4 = (1 + 0.035/12)^(12t) log 4 = 12t log(1.0029) t ≈ 39.666 years Doubling-Time Growth Formula If a population of size N₀ doubles every d units of time, then the number N in the population at time t is given by: N = N₀ * 2^(t/d) Half-Life Decay Formula If an amount N₀ has a half-life h, then the amount remaining at time t is: N = N₀ * (1/2)^(t/h) EXAMPLE 3 - Half-Life and Doubling Time a. The half-life of radioactive radon gas is 3.8 days. How much of 100 mg of the gas will be left after 7 days? N = 100(1/2)^(7/3.8) N ≈ 27.89 mg b. A 7 mg population of bacteria doubles every 53 weeks. Find the size of the population after 2 years. N = 7(2)^(104/53) N ≈ 27.28 mg