**OBJECTIVES**

1. Applications and Extensions of Logarithmic and Exponential Functions

2. Logistic Models

**Objective 1:** Applications and Extensions of Logarithmic and Exponential Functions

There are many occurrences in nature that can be modeled by exponential functions; exponential growth and decay.

Exponential word problems almost always take the growth/decay formula, y= ae^{kt}

- "y" is the ending amount of whatever you're dealing with
- "a" is the beginning amount of whatever you’re dealing with
*e*is euler's numbers- "k" is the growth or decay rate
- "t" is time

**Word Problems:**

Exponential Growth Example

(a.) 1,000 bacteria are growing continuously at a rate of 16% per day. Write a function that models the number of bacteria using the form, y=ae^{kt}

“a” is the starting amount→ 1000

“k” is the growth rate→ 16% or .16

“t” is the time→ t=days

“y” is the number of bacteria

y(t) is the number of bacteria after t, days

**y(t)=1000e ^{.16t}**

(b.) Using the model above, determine how many bacteria will be present after 30 days analytically

t=30 days

y=1000e^{.16(30)}

y=121,510.4175

After 30 days, there will be about 121,510 bacteria present.

(c.) Using the model, determine after how many days there will be 500,000 bacteria present

500,000=1000e^{.16t}

500=e^{.16t} *divide both sides by 1,000

ln500=lne^{.16t} *take ln of both sides

ln500=.16tln*e* *use “loga^{b}=bloga” property

1n500=.16t *lne is equal to 1

(ln500)/(.16)=(.16t)/(.16)

t=(ln500)/(.16)

t≈38.84130062

There will be 500,000 bacteria present after about 38.84 days.

**Exponential Decay Example**

(a.) The value V of a Porsche Cayenne that is t years old depreciates at a 18% per year. The starting price of the car when it was new was $60,000. Write an exponential function to model the situation in the form, y=ae^{-kt}

- “a” is the starting amount→ $60,000
- “k” is the decay rate→ 5% or -.5
- “t” is the time→ t=years
- “v” is the value of the car
- v(t) is value of the car (in dollars) after t, years

**v(t)=60,000e ^{-.05t}**

(b.) What is the value, v, of the car after 20 years?

t=20 years

v(t)=60,000e^{-.05(20)}

t=22,072.76647

According to the model, after 20 years, the car will be worth about $22,073.

(c.) Analytically determine after how many years the car will be worth $2,000

2000=60,000e^{-.05t}

1/30=e^{-.05t} *divide both sides by 60,000

ln(1/30)=lne^{-.05t} *take the ln of both sides

ln(1/30)=-.05tlne *use “loga^{b}=bloga” property

ln(1/30)=-.05t *lne is equal to 1

(ln(1/30))/(-.05)=(-.05t)(-.05)

t=(ln(1/30))/(-.05)

t≈68.02394763

The value of the car will be $2,000 after about 68 years.

**Real-World Application of Logarithmic Function**

(a.) The function r(x)=.56ln(x)+.0798 represents the running speed of a cheetah, in meters per second, of all the cheetah living in Africa, whose population, is x. Most other animals are slow compared to the cheetah's average pace of 29 meters per second. According to the model, what is the population of cheetahs in Africa?

29=.56ln(x)+24.996

4.004=.56ln(x) *subtract 24.996 from both sides

7.15=lne(x) *divide .56 both sides

e^{7.15}=x *ln is base e, change to exponential form

x=1,274.105955

If a cheetah’s average pace is 28 meters per second, than the population of cheetahs in Africa is around 1,274.

**Law of Uninhibited Growth and Decay**

The function of the form A(t)=A_{0}e^{kt} is an called an exponential growth function when k>0 and called an exponential decay function when k<0.

- A
_{0}and k are constants - The amount of A depends on the amount of t
- In this equation, A
_{0}is always equal to the value of “A” in the point (0,A) when the time is zero- Proof: Substitute (0,A) into the equation A(t)=A
_{0}e^{kt}- A(t)=A
_{0}e^{kt} - A=A
_{0}e^{k*0}*any number to the zero power is 1 - A=A
_{0}

- A(t)=A

- Proof: Substitute (0,A) into the equation A(t)=A

**Example of Exponential Growth:**

(a.) Suppose a bacteria population grows at a rate proportional to the population. The number of bacteria in the culture increases from 200 to 1,000 in 2 hours.

(i.) Two points that can generated from this problem situation are (0,200) and (2, 1000)

(1.) Of the form (t,A), t is time in hours passed and A is bacteria after t hours

(ii.) Using the two ordered pairs from (i), write a general exponential growth model of the form A(t)=A_{0}e^{kt}

200=A_{0}e^{k*0}

200=A_{0}e^{0} *any number to the zero power is one

A_{0}=200

1000=200e^{k*2} *substitute (2, 1000) into the model and the value for A_{0}

5=ek2 *divide both sides by 200

ln5=lne^{k2} *take the ln of both sides

ln5=k2lne *use properties of logs to bring down the exponent

ln5=2k *lne equals 1

(ln5)/2=(2k)/2 *divide both sides by 2 to isolate the variable

k=(ln5)/2

The model that represents this problem situation is A(t)=200e^{(ln5/2)t}

(iii.) Using the model found in part (ii), determine how long it will take the bacteria population to reach 20,000

A(t)=200e^{(ln5/2)t}

20,000=200e^{(ln5/2)t}

100=e^{(ln5/2)t} *divide both sides by 200

ln100=lne^{(ln5/2)t} *take the ln of both sides

ln100=((ln5/2)t)lne *use properties of logs to bring down the exponent

ln100=((ln5)/2)t

(ln100)/((ln5)/2)=(((ln5)/2)t)/((ln5)/2) *isolate the variable

t=(ln100)/((ln5)/2)

t≈5.722706

The bacteria population will reach 20,000 in about 5.7 hours.

**Example of Exponential Decay: Half-Life Problem**

(a.) Uranium-232 decays exponentially and has a half-life of approximately 70 years. Suppose there is a 100 kg sample of uranium-232

(i.) Two ordered pairs that lie on the decay function are (0,100) and (70, 50)

(1.) The ordered pair is of the form (t, A) where t is time in years and A is uranium-232 left after t years

(ii.) Substitute the ordered pairs into the model A(t)=A_{0}e^{kt} and write the resulting function

A_{0}=100

50=100e^{k*70} *substitute (70,50) and A_{0} into the model

½=e^{70k} *divide both sides by 100

ln½=lne^{70k} *take the ln of both sides

ln½=70klne *use properties of logs to bring down the exponent

ln½=70k *lne is equal to 1

(ln½)/70=(70k)/(70) *isolate the variable

k=(ln½)/70

The function that models this problem situation is A(t)=100e^{((ln½)/70)t}

(iii.) Using the model, find when 10 kg of uranium will be left

A(t)=100e^{((ln½)/70)t}

10=100e^{((ln½)/70)t} *substitute 10 in for “A”

1/10=e^{((ln½)/70)t} *divide both sides by 100

ln(1/10)=lne^{((ln½)/70)t} *take the ln of both sides

ln(1/10)=((ln½)/70)tlne *use properties of logs to bring down the exponent

ln(1/10)=(((ln½)/70)t) *lne is equal to 1

ln(1/10)/((ln½)/70)=(((ln½)/70)t)/((ln½)/70)*isolate the variable

t=ln(1/10)/((ln½)/70)

t≈232.53496664

According to the model, after about 232.5 years, 10 kg of uranium-232 will be left

**Objective 2:** Logistic Models

Exponential models represent when a population’s decay or growth is unconstrained. In reality, most populations have limiting factors that contribute to the size of the population. This limit is refereed to as the “carrying capacity”. Populations under these constrictions are best modeled by logistic functions.

- The logistic function is usually given the form P(t)=(c)/(1+ae
^{-bt})- “c” is the carrying capacity
- The carrying capacity is also the horizontal asymptote of the function

- “a” and “b” are constants with b>0
- “t” represents time
- “e” is euler’s number

- “c” is the carrying capacity

**Example of Logistic Growth:**

The number of tigers in the safari have many limiting factors that affect the population. Scientists working in Africa have been studying the population dynamics of the tiger population and have come to the conclusion that the growth of the tiger population in this particular safari is modeled by the logistic curve P(t)=(2000)/(1+3e^{-.145t}) where t is measured in months.

(a.) What is the carrying capacity (horizontal asymptote) of the number of tigers in the safari?

The carrying capacity is 2,000 tigers because that is the “c” value of the function. This is also the horizontal asymptote.

(b.) How many tigers were initially in the safari?

(1.) Substitute in t=0

(a) P(0)=(2000)/(1+3e^{-.145*0})

(b) P(0)=(2000)/(1+3e^{0}) *any number to the power of 0 is 1

(c) P(0)=2000/(1+3)

(d) P(0)=2000/4

(e) P(0)=500

There were initially 500 tigers in the safari.

(c.) When will the population of tigers be 1,500? Solve the equation analytically.

1500=(2000)/(1+3e^{-.145t})

1500/1=(2000)/(1+3e^{-.145t}) *make a proportion and cross multiply

2000=1500(1+3e^{-.145t}) *leave in factored form, don’t distribute

2000/1500=(1500(1+3e^{-.145t}))/1500 *divide by 1500

1.3333333=1+3e^{-.145t} *subtract 1 from both sides

.3333333=3e^{-.145t} *divide both sides by 3

.111111111=e^{-.145t} *take ln of both sides

ln(.11111111)=lne^{-.145t} *use properties of logs to bring down the exponent

ln(.11111111)=-.145tlne *lne is equal to 1

ln(.11111111)=-.145t *isolate the variable

(ln(.11111111))/(-.145)=(-.145t)/(-.145)

t=(ln(.11111111))/(-.145)

t≈15.15327295

The tiger population in the safari will be 1,500 in a little over 15 months.

**Practice Problems:**

1. Radioactive decay: An alien isotope of plutonium has just been found in the United States. It is said to have a half life of 238 years. If the entire substance is 8 kg, how much will be left after 100 years?

2. While studying an alien specimen you find that this carbon based life form residue there is 2.7g of Carbon-14 at 2pm, at 7pm there is 1.5 g left. What is the half-life of this substance

3. A new species of animal was discovered to initially have a population of 215. After 10 years of uninhibited growth, the population has grown to 1134. Write the exponential function in the form, A(t)=A_{0}e^{kt}, to model the situation. Then, find when the population will be 4,000.

4. Suppose that a species of fish in a lake is similar to a logistic population model. This lake has a growth rate of k = 0.3 per year. Additionally, it has a carrying capacity of c= 10000. If 2,500 fish are initially occupying the lake, analytically determine the function P(t) that models the lake after *t* years. Then use the model to find the number of fish after 8 years.