Transcript Exponential Growth and Decay

Exponential Growth and
Decay
Section 6.1
Given x = 2, y = ½, and z = 4.1
•
•
•
•
•
Evaluate each expression:
1.) 3y
2.) 42z
3.) 10(2)y+2
4.) -5yz
• You can use a calculator to model the
growth of 25 bacteria, assuming that
the entire population doubles every
hour.
Time
0
(hr)
Popula 25
tion
1
2
3
4
5
6
?
?
?
?
?
?
• Some things grow or decay at an
exponential rate rather than a steady
(linear) rate.
• This means they grow or decay very
rapidly.
• To find the amount after a certain time
period you must know 3 things:
– 1. original amount
– 2. growth/decay rate
– 3. Time period of growth/decay
• Use this basic expression to write an
expression for exponential growth/decay.
• (Original population)(growth)number of time periods
• Ex. The 25 bacteria in the original problem
doubles every hour, find the number of bacteria
after 10 hours
• Original population = 25
• Growth = 2 (since it doubles)
• Time periods = 10 (since it doubles each hour)
• Expression: 25 210
• Population after 10 hours = 25, 600
• Ex. 2 If 50 bacteria triple every 3 hours,
find the number of bacteria after 12
hours.
• Original amount = 50
• Growth = 3
• Time periods = 4 (every 3 hours it triples
12/3 = 4)
• Expression: 50 34
• Answer: 4050
• Ex. 3 200 bacteria double every 15
minutes, find the amount after 1.5 hours.
• Original amount: 200
• Growth: 2
• Time period: 6 (90 minutes/ 15 minutes)
• Expression: 200 26
• Answer: 12, 800
• Ex. 4 100 bacteria triple every 2
hours. Find the amount after 5
hours.
• Original = 100
• Growth = 3
• Time periods = 5/2
5
• Expression : 100 32
• Answer 1558.845… rounds to 1559
From yesterday,
• Use this basic expression to write an
expression for exponential growth/decay.
• (Original population)(growth)number of time periods
• Assuming an initial population of 100 bacteria,
predict the population of bacteria after n
hours if the population doubles.
• The population after n hours can be
represented by the following exponential
expression: 100 2n
• 100 2n is called an exponential expression
because the exponent, n, is a variable and the
base, 2, is a fixed number.
• The base of an exponential expression is called
the multiplier.
To find a multiplier
• Add or subtract the growth decay
rate from 100%
• Change to a decimal.
• Ex 1.
• 5.5% growth
• 100% + 5.5% = 105.5%
• Multiplier = 1.055
•
•
•
•
•
•
•
•
Ex. 2
0.25 growth
100% + 0.25% = 100.25%
Multiplier = 1.0025
Ex. 3
3% decay
100% - 3% = 97%
Multiplier = .97
•
•
•
•
Ex. 4
0.5% decay
100% - 0.5% = 99.5%
Multiplier = .995
Modeling Growth or Decay
• Ex. 1 Since 1990, the population of the United States
has been growing at a rate of 8% each decade. If the
population was 248,718,301 in 1990, predict the
number of people in 2020. Round to the nearest
hundred thousand.
• Original amount = 248,718,301
• Multiplier = 1.08 (100% + 8%)
• Time period = 3 (30 years/10 years)
• Expression: 248, 718,301 1.083
• Answer: 313,313,428.4
• 313, 300, 000 people
• Suppose you buy a car for $15,000.
Its value decreases at a rate of
about 8% per year. Predict the value
of the car after 4 years, and after 7
years.
• Solution:
• Multiplier: 100% - 8% = 92% = .92
n
• Exponential expression: 15,000  .92
• Value after 4 years: 15,000  .924  $10,745.89
• Value after 7 years: $8367.70
• Ex. 2 You invested $1000 in a company’s stock
at the end of 2009 and that stock has
increased at a rate of about 15% per year.
Predict the value of the stock, to the nearest
cent, at the end of the years 2014.
• Original amount: 1000
• Multiplier: 1.15 (100% + 15%)
• Time periods: 5 (5 years / 1 year)
• Expression: 1, 000 1.155
• Answer: $2011.36
• Ex. 3 You buy a new car for $15,000 and
its value decreases at a rate of about 8%
per year. Predict the value of the car, to
the nearest cent, after 4 years.
• Original amount: 15,000
• Multiplier: .92 (100% - 8%)
• Time: 4
• Expression: 15, 000 .924
• Answer: $10, 745.89