Exponential Decay

Exponential decay formula

Replace y with 0.5a and t with 2.6.

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.

Divide each side by a.

Exponential Decay

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.

Divide each side by –2.6.

In 0.5 = –2.6k Inverse Property of Exponents and Logarithms

In 0.5 = In e–2.6k Property of Equality for Logarithmic Functions

A. k = –0.0866

B. k = –4.1589

C. k = 0.0866

D. k = 4.1589

HEALTH The half-life of radioactive iodine used in medical studies is 8 hours. What is the value of k for radioactive iodine?

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 Formula for the decay of Sodium-22

0.1a = ae–0.2666t Replace y with 0.1a.

0.1 = e–0.2666t 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 andLogarithms

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.

A. about 0.05 hour later

B. about 0.39 hour later

C. about 2.58 hours later

D. about 18.58 hours later

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?

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

Use a calculator.

Answer: China’s relative rate of growth is about 0.0066, or about 0.66%.

Divide each side by 7.

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

Use a calculator.

Answer: China’s population will reach 1.5 billion in 2026.

ln e

x = xIn 1.1905 ≈ 0.0066t

Divide each side by 0.0066.

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

Use a calculator.

Answer: India’s population will surpass China’s in 18.5 years, or midway through 2025.

Product Property of Logarithms

In 1.32 + In e0.0066t < In 1.13 + In e0.015t

ln e

x = x In 1.32 + 0.0066t < In 1.13 + 0.015t

Subtract (0.0066t + ln 1.13) from each side.

In 1.32 – In 1.13 < 0.0084t

Divide each side by 0.0084.

Logistic Growth

Answer:

A. A city’s population in millions is modeled by

, where t is the number of years

since 2000. Graph the function.

Logistic Growth

Answer: The horizontal asymptote is at f (t) = 1.432.

B. A city’s population in millions is modeled by

, where t is the number of years

since 2000. What is the horizontal asymptote?

Logistic Growth

Answer: The population will reach a maximum of a little less than 1,432,000 people.

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?

Logistic Growth

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.

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?

A. B.

C. D.

A. A city’s population in millions is modeled by

where t is the number of years

since 2000. Graph the function.

A. f(t) = 2.971

B. f(t) = 1.13

C. f(t) = –0.28

D. f(t) = 1.563

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. 1 million people

B. 1.563 million people

C. 1.13 million people

D. 0.28 million people

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. by the year 2008

B. by the year 2010

C. by the year 2012

D. by the year 2014

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?

Homework: P. 513 # 1 - 11