It’s a Natural (Guided Practice)

The natural base, e, is an irrational number that approaches 2.71828…. An exponential function with base e is called a natural base exponential function, y = ex. Most calculators have an ex function for evaluating natural base expressions. Use a calculator to evaluate each expression to four decimal places. 1. e2 ≈ 7.3891

2. e−1.3 ≈ 0.2725

3. e5 ≈ 148.4132

4. e−2.2 ≈ 0.1108

The logarithm with base e is called the natural logarithm, sometimes denoted by loge x, but more often abbreviated ln x. The natural logarithmic function, y = ln x, is the inverse of the natural base exponential function, y = ex. Use a calculator to evaluate each expression to four decimal places. 1. ln 4 ≈ 1.3863

2. ln 0.05 ≈ −2.9957

3. ln 7 ≈ 1.9459

4. ln 0.25 ≈ −1.3863

Using the fact that ln x = loge x, write an equivalent exponential or logarithmic equation. (Remember: logb x = y if and only if by = x) 1. ex = 5 2. ln x ≈ 0.6931 3. ex = 6 4. ln x ≈ 0.5352

loge 5 = x ln 5 = x loge x ≈ 0.6931 x ≈ e0.6931 ln 6 = x x ≈ e0.5352

Since the natural base function and the natural logarithmic function are inverses, these two functions can be used to â€œundoâ€? each other.

Equations and inequalities involving base e are easier to solve using natural logarithms than using common logarithms. All the properties of logarithms that you have learned apply to natural logarithms as well.

Solve 5e−x −7 = 2 5e−x = 9

Add 7 to each side.

e−x = 9/5

Divide each side by 5.

ln e−x = ln 9/5

Property of Exponents and Logarithms.

−x = ln 9/5 x = −ln 9/5

Inverse Property of Exponents and Logarithms. Divide each side by −1.

x ≈ −0.5878 Use a calculator. Now check your answer by substituting −0.5878 into the original equation or by finding the intersection of the graphs of y = 5e−x −7 and y = 2.

Solve 5e−x −7 = 2 5e−x = 9

Add 7 to each side.

e−x = 9/5

Divide each side by 5.

ln e−x = ln 9/5

Property of Exponents and Logarithms.

−x = ln 9/5 x = −ln 9/5

Inverse Property of Exponents and Logarithms. Divide each side by −1.

x ≈ −0.5878 Use a calculator. Now check your answer by substituting −0.5878 into the original equation or by finding the intersection of the graphs of y = 5e−x −7 and y = 2.

Solve 5e−x −7 = 2 5e−x = 9

Add 7 to each side.

e−x = 9/5

Divide each side by 5.

ln e−x = ln 9/5

Property of Exponents and Logarithms.

−x = ln 9/5 x = −ln 9/5

Inverse Property of Exponents and Logarithms. Divide each side by −1.

x ≈ −0.5878 Use a calculator. Now check your answer by substituting −0.5878 into the original equation or by finding the intersection of the graphs of y = 5e−x −7 and y = 2.

Solve 5e−x −7 = 2 5e−x = 9

Add 7 to each side.

e−x = 9/5

Divide each side by 5.

ln e−x = ln 9/5

Property of Exponents and Logarithms.

−x = ln 9/5 x = −ln 9/5

Inverse Property of Exponents and Logarithms. Divide each side by −1.

Solve 5e−x −7 = 2 5e−x = 9

Add 7 to each side.

e−x = 9/5

Divide each side by 5.

ln e−x = ln 9/5

Property of Exponents and Logarithms.

−x = ln 9/5 x = −ln 9/5

Inverse Property of Exponents and Logarithms. Divide each side by −1.

Solve 5e−x −7 = 2 5e−x = 9

Add 7 to each side.

e−x = 9/5

Divide each side by 5.

ln e−x = ln 9/5

Property of Exponents and Logarithms.

−x = ln 9/5 x = −ln 9/5

Inverse Property of Exponents and Logarithms. Divide each side by −1.

Solve 5e−x −7 = 2 5e−x = 9

Add 7 to each side.

e−x = 9/5

Divide each side by 5.

ln e−x = ln 9/5

Property of Exponents and Logarithms.

−x = ln 9/5 x = −ln 9/5

Inverse Property of Exponents and Logarithms. Divide each side by −1.

Solve each equation. 1. 3ex + 2 = 4

−0.4055

2. 4e−x − 9 = −2

−0.4055

Growing or Decaying? (Guided Practice)

When a quantity increases by a fixed percent each time period, the amount y of that quantity after t time periods is given by y = a(1 + r)t, where a is the initial amount and r is the percent of increase expressed as a decimal. The percent of increase r is also referred to as the rate of growth.

1) In 1910, the population of a city was 120,000. Since then, the population has increased by 1.5% per year. If the population continues to grow at this rate, what will the population be in 2010? Read the problem carefully. You need to find the population of the city 2010 â€“ 1910, or 100, years later. Since the population is growing at a fixed percent each year, use the formula y = a(1 + r)t .

Solve the problem. y = a(1 + r)t

Exponential growth formula

= 120,000(1 + 0.015)100

Replace a with 120,000, r with 0.015, and t with 100.

= 120,000(1.015)100

Simplify.

â‰ˆ 531,845

Use a calculator.

Solve the problem. y = a(1 + r)t

Exponential growth formula

= 120,000(1 + 0.015)100

Replace a with 120,000, r with 0.015, and t with 100.

= 120,000(1.015)100

Simplify.

â‰ˆ 531,845

Use a calculator.

Solve the problem. y = a(1 + r)t

Exponential growth formula

= 120,000(1 + 0.015)100

Replace a with 120,000, r with 0.015, and t with 100.

= 120,000(1.015)100

Simplify.

â‰ˆ 531,845

Use a calculator.

Solve the problem. y = a(1 + r)t

Exponential growth formula

= 120,000(1 + 0.015)100

Replace a with 120,000, r with 0.015, and t with 100.

= 120,000(1.015)100

Simplify.

â‰ˆ 531,845

Use a calculator.

2) Home values in Millersport increase about 4% per year. Mr. Thomas purchased his home eight years ago for $122,000. What is the value of his home now? $166,965

Another model for exponential growth, preferred by scientists, is y = aekt, where k is a constant

3) As of 2005, China was the world’s most populous country, with an estimated population of 1.31 billion people. The second most populous country was India, with 1.08 billion. The populations of India and China can be modeled by I(t) = 1.08e0.0103t and C(t) = 1.31e0.0038t, respectively. According to these models, when will India’s population be more than China’s? Find t such that I(t) > C(t) I(t) > C(t) 1.08e0.0103t > 1.31 e0.0038t

Replace I(t) with 1.08e0.0103t and C(t) with 1.31e0.0038t

3) As of 2005, China was the world’s most populous country, with an estimated population of 1.31 billion people. The second most populous country was India, with 1.08 billion. The populations of India and China can be modeled by I(t) = 1.08e0.0103t and C(t) = 1.31e0.0038t, respectively. According to these models, when will India’s population be more than China’s? Find t such that I(t) > C(t) I(t) > C(t) 1.08e0.0103t > 1.31 e0.0038t

Replace I(t) with 1.08e0.0103t and C(t) with 1.31e0.0038t

ln 1.08e0.0103t > ln 1.31e0.0038t

Property of Inequality for Logarithms

ln 1.08 + ln e0.0103t > ln 1.31 + ln e0.0038t Product Property of Logarithms ln 1.08 + 0.0103t > ln 1.31 + 0.0038t Inverse Property of Exponents and Logarithms

ln 1.08e0.0103t > ln 1.31e0.0038t

Property of Inequality for Logarithms

ln 1.08 + ln e0.0103t > ln 1.31 + ln e0.0038t Product Property of Logarithms ln 1.08 + 0.0103t > ln 1.31 + 0.0038t Inverse Property of Exponents and Logarithms

ln 1.08e0.0103t > ln 1.31e0.0038t

Property of Inequality for Logarithms

ln 1.08 + ln e0.0103t > ln 1.31 + ln e0.0038t Product Property of Logarithms ln 1.08 + 0.0103t > ln 1.31 + 0.0038t Inverse Property of Exponents and Logarithms

0.065t > ln 1.31 â€“ ln 1.08

Subtract 0.0038t from each side.

ln1.31 - ln1.08 t> 0.0065

Divide each side by 0.006.

t > 29.70

Use a calculator.

After 30 years, or in 2035, India will be the most populous country in the world.

0.065t > ln 1.31 â€“ ln 1.08

Subtract 0.0038t from each side.

ln1.31 - ln1.08 t> 0.0065

Divide each side by 0.006.

t > 29.70

Use a calculator.

After 30 years, or in 2035, India will be the most populous country in the world.

0.065t > ln 1.31 â€“ ln 1.08

Subtract 0.0038t from each side.

ln1.31 - ln1.08 t> 0.0065

Divide each side by 0.006.

t > 29.70

Use a calculator.

After 30 years, or in 2035, India will be the most populous country in the world.

4) Two different types of bacteria in two different cultures reproduce exponentially. The first type can be modeled by B1(t) = 1200e0.1532t, and the second can be modeled by B2(t) = 3000 e0.0466t. According to these models, how many hours will it take for the amount of B1 to exceed the amount of B2? 8.60 hours

When a quantity decreases by a fixed percent each year, or other period of time, the amount y of that quantity after t years is given by y = a(1 â€“ r)t, where a is the initial amount and r is the percent of decrease expressed as a decimal. The percent of decrease r is also referred to as the rate of decay. 5) A cup of coffee contains 130 milligrams of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated?

Read the problem carefully. The problem gives the amount of caffeine consumed and the rate at which the caffeine is eliminated. It asks you to find the time it will take for half of the caffeine to be eliminated. Use the formula y = a(1 â€“ r)t. Let t be the number of hours since drinking the coffee. The amount remaining y is half of 130 or 65.

Solve the problem. y = a(1 â€“ r)t

Exponential decay formula

65 = 130(1 â€“ 0.11)t Replace y with 65, a with 130, and r with 0.11. 0.5 = (0.89)t

Divide each side by 130 and simplify.

log 0.5 = log (0.89)t

Property of Equality for Logarithms

log 0.5 = t log (0.89)

Power Property for Logarithms

Solve the problem. y = a(1 â€“ r)t

Exponential decay formula

65 = 130(1 â€“ 0.11)t Replace y with 65, a with 130, and r with 0.11. 0.5 = (0.89)t

Divide each side by 130 and simplify.

log 0.5 = log (0.89)t

Property of Equality for Logarithms

log 0.5 = t log (0.89)

Power Property for Logarithms

Solve the problem. y = a(1 â€“ r)t

Exponential decay formula

65 = 130(1 â€“ 0.11)t Replace y with 65, a with 130, and r with 0.11. 0.5 = (0.89)t

Divide each side by 130 and simplify.

log 0.5 = log (0.89)t

Property of Equality for Logarithms

log 0.5 = t log (0.89)

Power Property for Logarithms

Solve the problem. y = a(1 â€“ r)t

Exponential decay formula

65 = 130(1 â€“ 0.11)t Replace y with 65, a with 130, and r with 0.11. 0.5 = (0.89)t

Divide each side by 130 and simplify.

log 0.5 = log (0.89)t

Property of Equality for Logarithms

log 0.5 = t log (0.89)

Power Property for Logarithms

Solve the problem. y = a(1 â€“ r)t

Exponential decay formula

65 = 130(1 â€“ 0.11)t Replace y with 65, a with 130, and r with 0.11. 0.5 = (0.89)t

Divide each side by 130 and simplify.

log 0.5 = log (0.89)t

Property of Equality for Logarithms

log 0.5 = t log (0.89)

Power Property for Logarithms

log 0.5 =t log 0.89 5.9480 â‰ˆ t

Divide each side by log 0.89.

Use a calculator.

It will take approximately 6 hours.

log 0.5 =t log 0.89 5.9480 â‰ˆ t

Divide each side by log 0.89.

Use a calculator.

It will take approximately 6 hours.

6) A store is offering a clearance sale on a certain type of digital camera. The original price for the camera was $198. The price decreases 10% each week until all of the cameras are sold. How many weeks will it take for the price of the cameras to drop below half of the original price? 7 weeks

Another model for exponential decay is given by y = aeâˆ’kt, where k is a constant. This is the model preferred by scientist. Use this model to solve problems involving radioactive decay. Radioactive decay is the decrease in the intensity of a radioactive material over time. Being able to solve problems involving radioactive decay allows scientists to use carbon dating methods.

7) The half-life of a radioactive substance is the time it takes for half of the atoms of the substance to disintegrate. All life on Earth contains Carbon-14, which decays continuously at a fixed rate. The half-life of Carbon-14 is 5760 years. That is, every 5760 years half of a mass of Carbon-14 decays away. What is the value of k and the equation of decay for Carbon-14? Let a be the initial amount of the substance. The amount y that remains after 5760 years is then represented by 0.5a.

y = ae−kt

Exponential decay formula

0.5a = ae−k(5760)

Replace y with 0.5a and t with 5760.

0.5 = e−5760k

Divide each side by a.

ln 0.5 = ln e−5760k

Property of Equality for Logarithmic Functions

ln 0.5 = −5760k

Inverse Property of Exponents and Logarithms

ln 0.5 =k - 5760

Divide each side by −5760.

y = ae−kt

Exponential decay formula

0.5a = ae−k(5760)

Replace y with 0.5a and t with 5760.

0.5 = e−5760k

Divide each side by a.

ln 0.5 = ln e−5760k

Property of Equality for Logarithmic Functions

ln 0.5 = −5760k

Inverse Property of Exponents and Logarithms

ln 0.5 =k - 5760

Divide each side by −5760.

y = ae−kt

Exponential decay formula

0.5a = ae−k(5760)

Replace y with 0.5a and t with 5760.

0.5 = e−5760k

Divide each side by a.

ln 0.5 = ln e−5760k

Property of Equality for Logarithmic Functions

ln 0.5 = −5760k

Inverse Property of Exponents and Logarithms

ln 0.5 =k - 5760

Divide each side by −5760.

y = ae−kt

Exponential decay formula

0.5a = ae−k(5760)

Replace y with 0.5a and t with 5760.

0.5 = e−5760k

Divide each side by a.

ln 0.5 = ln e−5760k

Property of Equality for Logarithmic Functions

ln 0.5 = −5760k

Inverse Property of Exponents and Logarithms

ln 0.5 =k - 5760

Divide each side by −5760.

y = ae−kt

Exponential decay formula

0.5a = ae−k(5760)

Replace y with 0.5a and t with 5760.

0.5 = e−5760k

Divide each side by a.

ln 0.5 = ln e−5760k

Property of Equality for Logarithmic Functions

ln 0.5 = −5760k

Inverse Property of Exponents and Logarithms

ln 0.5 =k - 5760

Divide each side by −5760.

y = ae−kt

Exponential decay formula

0.5a = ae−k(5760)

Replace y with 0.5a and t with 5760.

0.5 = e−5760k

Divide each side by a.

ln 0.5 = ln e−5760k

Property of Equality for Logarithmic Functions

ln 0.5 = −5760k

Inverse Property of Exponents and Logarithms

ln 0.5 =k - 5760

Divide each side by −5760.

- 0.6931472 ≈k - 5760 0.00012 ≈ k

Use a calculator. Simplify.

The value of k for Carbon-14 is 0.00012. Thus, the equation for the decay of Carbon-14 is y = ae−0.00012t, where t is given in years.

- 0.6931472 ≈k - 5760 0.00012 ≈ k

Use a calculator. Simplify.

The value of k for Carbon-14 is 0.00012. Thus, the equation for the decay of Carbon-14 is y = ae−0.00012t, where t is given in years.

8) A paleontologist examining the bones of a woolly mammoth estimates that they contain only 3% as much Carbon-14 as they would have contained when the animal was alive. How long ago did the mammoth die? Let a be the initial amount of Carbon-14 in the animalâ€™s body. Then the amount y that remains after t years is 3% of a or 0.03a.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

Property of Equality for Logarithms Inverse Property of Exponents and Logarithms Divide each side by −0.00012 Use a calculator.

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

Property of Equality for Logarithms Inverse Property of Exponents and Logarithms Divide each side by −0.00012 Use a calculator.

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

Property of Equality for Logarithms Inverse Property of Exponents and Logarithms Divide each side by −0.00012 Use a calculator.

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

The mammoth lived about 29,000 years ago.

y = ae−0.00012t

Formula for the decay of Carbon-14

0.03a = ae−0.00012t

Replace y with 0.03a.

0.03 = e−0.00012t

Divide both sides by a.

ln 0.03 = ln e−0.00012t ln 0.03 = −0.00012t ln 0.03 =t - 0.00012

29,221 ≈ t

The mammoth lived about 29,000 years ago.

9) A specimen that originally contained 150 milligrams of Carbon-14 now contains 130 milligrams. How old is the fossil? About 1193 years

(Homework)

3. A piece of machinery valued at $250,000 depreciates a fixed rate of Name _________________ Periodat ___ Date _________ 12% per year. After how many years will the value have depreciated to $100,000?

1.5.Alex invests $2000 an account at that a 6% rate annual rateper of month. growth.IfTo A computer systemindepreciates anhas average of 4% the when will system the investment be worth $3600?in how many months was originally $12,000, thenearest value ofyear, the computer worth $7350?

4. The value of a stock when purchased was $10 a share. The stock grew at a monthly rate of 7%. Find the number of months it took to reach a value of $15.75 a share.

2. The value of a stock when purchased is $10 a share. The stock decreased at of 3% daily. Find the number of days it took for the stock to be worth $7.40.

6. How many days will it take a culture of bacteria to increase from 2000 to The Ares Curriculum Project, Exponential and Logarithmic Relations, Adapted from Glencoe Algebra 2, 110507_hw_growing_growing_growing, page 1 50,000 if the growth rate per day is 93.2%? of 2

3. A piece of machinery valued at $250,000 depreciates at a fixed rate of

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