Exponential Decay
GEOLOGY The half-life of Sodium-22 is 2.6 years.
Determine the value of k and the equation of decay
for Sodium-22.
If a is the initial amount of the substance, then the
amount y that remains after 2.6 years is
or 0.5a.
Exponential decay formula
Replace y with 0.5a and t with 2.6.
Divide each side by a.
Exponential Decay
In 0.5 = In e–2.6k
In 0.5 = –2.6k
Property of Equality for Logarithmic
Functions
Inverse Property of Exponents and
Logarithms
Divide each side by –2.6.
0.2666 ≈ k
Use a calculator.
Answer: The value of k of Sodium-22 is 0.2666. Thus,
the equation for the decay of Sodium-22 is
y = ae–0.2666t, where t is given in years.
HEALTH The half-life of radioactive iodine used in
medical studies is 8 hours. What is the value of k
for radioactive iodine?
A. k = –0.0866
B. k = –4.1589
C. k = 0.0866
D. k = 4.1589
Carbon Dating
GEOLOGY A geologist examining a meteorite
estimates that it contains only about 10% as much
Sodium-22 as it would have contained when it
reached the surface of the Earth. How long ago did
the meteorite reach the surface of the Earth?
Understand
The formula for the decay of Sodium-22 is y = ae–kt. You
want to find how long ago the meteorite reached Earth.
Plan
Let a be the initial amount of Sodium-22 in the meteorite.
The amount y that remains after t years is 10% of a or
0.10a.
Carbon Dating
Solve
y = ae–0.2666t
0.1a = ae–0.2666t
0.1 = e–0.2666t
Formula for the decay of Sodium-22
Replace y with 0.1a.
Divide each side by a.
In 0.1= ln e–0.2666t
Property of Equality for Logarithms
ln 0.1= –0.2666t
Inverse Property for Exponents and
Logarithms
Divide each side by –0.2666.
8.64 ≈ t
Use a calculator.
Carbon Dating
Answer: It reached the surface of Earth about 8.6
years ago.
Check
Use the formula to find the amount of the sample
remaining after 8.6 years. Use an original amount of 1.
y = ae-0.2666t
Original equation
= 1e-0.2666(8.6)
a = 1 and t = 8.6
≈ 0.101 or 10%
Use a calculator.
HEALTH The half-life of radioactive iodine used in
medical studies is 8 hours. A doctor wants to know
when the amount of radioactive iodine in a patient’s
body is 20% of the original amount. When will this
occur?
A. about 0.05 hour later
B. about 0.39 hour later
C. about 2.58 hours later
D. about 18.58 hours later
Continuous Exponential Growth
A. POPULATION In 2007, the population of China
was 1.32 billion. In 2000, it was 1.26 billion.
Determine the value of k, China’s relative rate of
growth.
Formula for continuous
exponential growth
y = 1.32, a = 1.26, and
t = 2007 – 2000 or 7
Divide each side by 1.26.
Continuous Exponential Growth
Property of Equality for
Logarithmic Equations
ln e x = x
Divide each side by 7.
Use a calculator.
Answer: China’s relative rate of growth is about
0.0066, or about 0.66%.
Continuous Exponential Growth
B. POPULATION In 2007, the population of China
was 1.32 billion. In 2000, it was 1.26 billion. When
will China’s population reach 1.5 billion?
Formula for continuous
exponential growth
y = 1.5, a = 1.26, and
k = 0.0066
≈
Divide each side by 1.26.
≈
Property of Equality for
Logarithmic Functions
Continuous Exponential Growth
In 1.1905 ≈ 0.0066t
ln e x = x
Divide each side by 0.0066.
Use a calculator.
Answer: China’s population will reach 1.5 billion
in 2026.
Continuous Exponential Growth
C. POPULATION In 2007, the population of China
was 1.32 billion. India’s population in 2007 was
1.13 billion and can be modeled by y = 1.13e0.015t.
Determine when India’s population will surpass
China’s. (Note: t represents years after 2007.)
Formula for
exponential
growth
Property of
Inequality for
Logarithms
Continuous Exponential Growth
In 1.32 + In e0.0066t < In 1.13 + In e0.015t
In 1.32 + 0.0066t < In 1.13 + 0.015t
In 1.32 – In 1.13 < 0.0084t
Product Property
of Logarithms
ln e x = x
Subtract
(0.0066t + ln 1.13)
from each side.
Divide each side
by 0.0084.
Use a calculator.
Answer: India’s population will surpass China’s in
18.5 years, or midway through 2025.
Logistic Growth
A. A city’s population in millions is modeled by
, where t is the number of years
since 2000. Graph the function.
Answer:
Logistic Growth
B. A city’s population in millions is modeled by
, where t is the number of years
since 2000. What is the horizontal asymptote?
Answer: The horizontal asymptote is at f(t) = 1.432.
Logistic Growth
C. A city’s population in millions is modeled by
, where t is the number of years
since 2000. What will be the maximum population?
Answer: The population will reach a maximum of a
little less than 1,432,000 people.
Logistic Growth
D. A city’s population in millions is modeled by
, where t is the number of years
since 2000. According to the function, when will the
city’s population reach 1 million?
Answer: The graph indicates the population will reach
1 million people at t ≈ 3. Replacing f(t) with
1 and solving for t in the equation yields
t = 2.78 years. So, the population of the city
will reach 1 million people by 2003.
A. A city’s population in millions is modeled by
where t is the number of years
since 2000. Graph the function.
A.
B.
C.
D.
B. A city’s population in millions is modeled by
where t is the number of years
since 2000. What is the horizontal asymptote?
A. f(t) = 2.971
B. f(t) = 1.13
C. f(t) = –0.28
D. f(t) = 1.563
C. A city’s population in millions is modeled by
where t is the number of years
since 2000. What will be the maximum population?
A. 1 million people
B. 1.563 million people
C. 1.13 million people
D. 0.28 million people
D. A city’s population in millions is modeled by
where t is the number of years
since 2000. According to the function, when will the
city’s population reach 1.5 million?
A. by the year 2008
B. by the year 2010
C. by the year 2012
D. by the year 2014
Homework: P. 513 # 1 - 11