Various Forms of Exponential Functions Doubling M is the total number after time t. c is the initial amount at time t = 0. t is the time d is the doubling period Half-life M is the total number after time t. c is the initial amount at time t = 0. t is the time h is the half-life

A = P(1 + i)n 1 Compound Interest A is the final amount including interest and principal A = P(1 + i)n P is the original principal invested i is the interest rate per compounding period n is the number of compounding periods 1

A(t) = c(a)x Exponential Function (General Form) A is the total amount or number (at time t) c is the initial amount or number a is the growth factor or decay rate x is the number of growth or decay periods

Applications of Exponential Functions Example 1 Two people are sent an email at 12 p.m. At 1:00 p.m. they send the email to 3 people. At 2:00 p.m. they send the email to each of 3 people. At 3:00 p.m. they send the email to each of 3 people. etc, etc. hour 1 2 3 4 5 6 number of people 2 6 18 54 162 486 1458

Equivalent expression hour 1 2 3 4 5 6 number of people 2 6 18 54 162 486 1458 Hour People told Equivalent expression 1 2 3 4 5 6 2 2 2×3 6 2×32 18 54 2×33 162 2×34 486 2×35 1458 2×36 N = 2×3t

A(t) = c(a)t , a > 1 Exponential Growth N = c(a)t c = initial amount a = growth rate Example 2: A bacteria dish begins with 50 bacteria. If the doubling period is 1 hour, how many bacteria are there after (i) 5 hours? (ii) 10 hours? N = c(a)t (i) N = 50(2)t (ii) N = 50(2)10 = 50(2)5 = 50(1024) = 51 200 bacteria = 50(32) = 1600 bacteria

A(t) = c(a)t = 25(2)5 = 25(32) = 800 bacteria Example 3: A bacteria dish begins with 25 bacteria. If the doubling period is 2 hours, how many bacteria are there after 10 hours? Solution 1: There are five doubling periods in 10 hours A(t) = c(a)t = 25(2)5 = 25(32) = 800 bacteria

N = c(a)t 2 = 1(a)2 2 = a2 = 800 bacteria Example (continued): A bacteria dish begins with 25 bacteria. If the doubling period is 2 hours, how many bacteria are there after 10 hours? Solution 2: If you start with one bacteria, there are 2 after 2 hours. N = c(a)t 2 = 1(a)2 2 = a2 = 800 bacteria

Determining the Growth Rate Example 4: In the year 2000, the population of a town was 28 090 people. In 2002, the population was 31 562. What is the growth rate? What was the population in 1998? Let 1998 be the initial time period when t = 0. A(t)= c(a)t A(t) = c(a)t 28090 = c(1.06)2 28090 = c(a)2 1.1236 = a2 31562 = c(a)4 (divide to find a) 25000 = c 1.06 = a

The truck depreciates in value by 20% per year. Exponential Decay A(t) = c(a)t , a < 1 c = initial amount a = decay rate Example 5: After 5 years, a $64 000 truck is worth $20 971.52. What is the decay rate? A(t) = c(a)t 20971.52 = 64000(a)5 0.80 = a The truck depreciates in value by 20% per year. 0.32768 = a5

Ex. 6 The half-life of a substance is 4 hours Ex.6 The half-life of a substance is 4 hours. How much of the substance will remain after (i) 12 hours (ii) 2 days if there were 64 grams at the start? (i) (ii) M = 0.0156 grams M = 8 grams

Ex 7: A bacteria dish begins with 100 bacteria Ex 7: A bacteria dish begins with 100 bacteria. After 5 hours there are 1600 bacteria. (i) What is the doubling period? (ii) Find the population after t hours. (iii) Determine the number of bacteria after 8 h. (i) the doubling period is 1.25 hours.

Ex 7(con’d): A bacteria dish begins with 100 bacteria Ex 7(con’d): A bacteria dish begins with 100 bacteria. After 5 hours there is 1600 bacteria. (i) What is the doubling period? (ii) Find the population after t hours. (iii) Determine the number of bacteria after 8 h. (iii) (ii) M = 100(2)6.4 = 100(84.44) = 8444 After 8 hours there will be 8444 bacteria.