MAT 124 Exponential & Logarithmic Models

. Exponential Growth Models

Here an amount 퐴 is growing at an uninhibited rate according to the following law:

푘푡 퐴(푡) = 퐴0푒 , where 푡 is the time elapsed (measured in seconds, minutes, hours, days, years, etc.),

퐴0 is the original amount present (grams, ounces, price, population, etc.), 푘 > 0 is the growth factor to be determined by the data given in the problem.

Example 1: World Population

In 2015, the world population was 7.21 billion people and was growing at a rate of 1.1% per year. a) Find an exponential growth model for world population. b) According to this model, what will be the world’s population in the year 2025? c) According to this model, in what year will the world population reach 10 billion people?

Solution

풌풕 풌풕 a) An exponential growth model for world population is given by 푷(풕) = 푷ퟎ풆 = ퟕ. ퟐퟏ풆 , where 푃(푡) is the world population, in billions, 푡 years after 2015.

To determine the growth factor 푘, we use the fact that the world population grows to 7.21(1.011) = 7.28931 billion in 2016. Thus,

7.28931 푃(1) = 7.21푒푘 = 7.28931 ⇒ 푒푘 = = 1.011 ⇒ 풌 = 퐥퐧(ퟏ. ퟎퟏퟏ) 7.21

We therefore have

푷(풕) = ퟕ. ퟐퟏ풆퐥퐧(ퟏ.ퟎퟏퟏ)풕 = ퟕ. ퟐퟏ(ퟏ. ퟎퟏퟏ)풕

since 푒ln(1.011) = 1.011. b) Since 푃(10) = 7.21(1.011)10 ≈ 8.044, we conclude that, according to this model, the world population is projected to growth to approximately 8.044 billion by 2025. c) Solving the equation 푃(푡) = 10 yields

10 10 7.21푒ln(1.011)푡 = 10 ⇒ 푒ln(1.011)푡 = ⇒ ln(1.011) 푡 = ln ( ) 7.21 7.21 ퟏퟎ 퐥퐧 ( ) ⇒ 풕 = ퟕ. ퟐퟏ ≈ ퟐퟗ. ퟗ 퐥퐧(ퟏ. ퟎퟏퟏ)

Therefore, the model predicts that the world population will reach 10 billion by the end of the year 2044. [See graph of 푃(푡) below.]

. Exponential Decay Models

Here an amount 퐴 is decaying exponentially according to the following law:

−푘푡 퐴(푡) = 퐴0푒 , where 푡 is the time elapsed (measured in seconds, minutes, hours, days, years, etc.),

퐴0 is the original amount present (grams, ounces, price, population, etc.), −푘 < 0 is the decay factor to be determined by the data given in the problem.

Example 2: Estimating the Age of an Ancient Bone

A bone fragment found in an archaeological dig contains 20% of its original Carbon-14. Knowing that the half-life of Carbon-14 is 5,730 years, how old is this bone? Round to the nearest year.

Solution

Suppose the original amount of Carbon-14 present in the bone was 퐴0. Then, since the half-life of Carbon-14 is 5,730 years, we determine 푘 as follows:

퐴 1 1 퐥퐧 ퟐ 퐴 푒−5730푘 = 0 ⇒ 푒−5730푘 = ⇒ −5730푘 = ln ( ) = − ln 2 ⇒ 풌 = 0 2 2 2 ퟓퟕퟑퟎ

ln 2 − 푡 So the amount of Carbon-14 in the bone is given by 퐴(푡) = 퐴0푒 5730 .

Solving the equation 퐴(푡) = .2퐴0 (since 20% of 퐴0 = .2퐴0) yields

ln 2 ln 2 − 푡 − 푡 ln 2 1 퐴 푒 5730 = .2퐴 ⇒ 푒 5730 = .2 ⇒ − 푡 = ln(. 2) = ln ( ) = − ln 5 0 0 5730 5 −ln 5 ퟓퟕퟑퟎ ∙ 퐥퐧 ퟓ ⇒ 푡 = ⇒ 풕 = ≈ ퟏퟑ, ퟑퟎퟓ ln 2 − 퐥퐧 ퟐ 5730

We conclude that this ancient bone is, approximately, 13,305 years-old.

See graph of 퐴(푡) below with 퐴0 = 1.

. Logistic Growth Models

This model is much more realistic than the previous exponential growth model. Here an amount 푃 grows according to the following logistic law:

푐 푃(푡) = , 1 + 푎푒−푏푡 where 푡 is the time elapsed (measured in seconds, minutes, hours, days, years, etc.), 푐 is the initial amount present (grams, ounces, price, population, etc.), 1+푎 푐 > 0 is the carrying capacity, or limiting value, of the model (i.e. as 푡 → ∞, 푃(푡) → 푐), 푏 > 0 is a constant determined by the rate of growth. Example 3: Koi Population

The population 푃(푡) of fish in a koi pond after 푡 months is modeled by the logistic function

68 푃(푡) = 1 + 16푒−0.28푡 a) What was the initial population of koi in the pond? b) How many koi will be in the pond after one and a half years? c) According to this model, when will the population of koi reach 65? Round to the nearest month. d) What happens to the koi population in the long term (after, say, 30 years)?

Solution

68 68 a) There were 4 koi in the pond initially since 푃(0) = = = 4. 1+16 17 b) There will be 61 koi in the pond after one and a half years since 푃(18) ≈ 61.62. c) Solving the equation 푃(푡) = 65 yields

68 = 65 ⇒ 68 = 65(1 + 16푒−0.28푡) = 65 + 1040푒−0.28푡 1 + 16푒−0.28푡 3 ⇒ 1040푒−0.28푡 = 68 − 65 = 3 ⇒ 푒−0.28푡 = 1040 ퟑ 3 퐥퐧 ( ) ⇒ −0.28푡 = ln ( ) ⇒ 풕 = − ퟏퟎퟒퟎ ≈ ퟐퟎ. ퟖퟖퟕ 1040 ퟎ. ퟐퟖ

Therefore, the model predicts that the koi population in the pond reaches 65 after 20 months, or one year and 8 months. d) In the long term, the koi population will approach 68. This is the carrying capacity of the model.

See the graph of 푃(푡) below.