PROBLEMS ON EXPONENTIAL FUNCTION CONTEXT AND DATA MODELING

For each problem, create a function to model the scenario.

Problem 1 :

A population 𝑝 of 500 people doubles every 35 years 𝑡.

Solution :

a = 500

Doubles every 35 years. So, d = 35

f(t) = a(2)^t/d 

f(t) = 500 (2)^t/35 

Problem 2 :

Mr. Kelly bought a new tractor for his farm in New York. It cost him $150,000. Unfortunately, it’s value 𝑣 depreciates in value by 5.4% per year 𝑡.

Solution :

Exponential decay :

v(t) = p(1 - r%)^t

p = 150000, r = 5.4%

v(t) = 150000(1 - 5.4%)^t

v(t) = 150000(1 - 0.054)^t

= 150000 (0.946)^t

Problem 3 :

A baseball card is worth $50 and its value 𝑣 increases at a rate of 23.5% per year 𝑡

Solution :

Exponential growth :

v(t) = p(1 + r%)^t

p = 50, r = 23.5%

v(t) = 50(1 + 23.5%)^t

= 50(1 + 0.235)^t

= 50(1.235)^t

Problem 4 :

There is 500 grams 𝑔 of radioactive material. Its halflife is 5,700 years, 𝑡

Solution :

A = A₀ (1/2) ^t/h

Here A₀ = 500, h = 5700

A = 500 (1/2) ^t/5700

Problem 5 :

700 grams of radioactive material 𝑚 decays at a rate of 2.4% per year 𝑡. How much material will be there after 100 years ?

Solution :

Exponential decay :

m(t) = p(1 - r%)^t

p = 700, r = 2.4%

m(t) = 700(1 - 2.4%)^t

= 700(1 - 0.024)^t

m(t) = 700(0.976)^t

The quantity of material after 100 years :

m(100) = 700(0.976)¹⁰⁰

= 61.67 grams.

So, after 100 years 61.67 grams of material will be there.

Problem 6 :

The new tires on a truck have a tread depth of 0.5 inches and decays at the rate of 1.6% per week. How deep will the tread be after 52 weeks ?

Solution :

Exponential decay :

m(t) = p(1 - r%)^t

p = 0.5, r = 1.6%

m(t) = 0.5(1 - 1.6%)^t

= 0.5(1 - 0.016)^t

m(t) = 0.5(0.984)^t

The deep of tread after 52 weeks :

t = 52

m(52) = 0.5(0.984)⁵²

= 0.216

Approximately 0.22 inches

Problem 7 :

A car that is worth $25000, decreases in value by 15% per year. How much will the car be after 5 years ?

Solution :

Exponential decay :

v(t) = p(1 - r%)^t

p = 25000, r = 15%

v(t) = 25000(1 - 15%)^t

= 25000(1 - 0.15)^t

= 25000(0.85)^t

Value after 5 years :

When t = 5

v(5) = 25000(0.85)⁵

= 25000(0.4437)

= 11092.63

So, the worth of car after 5 years is $11092.63.

Problem 8 :

Mr. Brust IQ currently 173, but it is decaying at a rate of 4.5% every year. What will Mr. Brust's IQ be in 20 years ?

Solution :

Exponential decay :

v(t) = p(1 - r%)^t

p = 173, r = 4.5 %

v(t) = 173(1 - 4.5%)^t

= 173(1 - 0.045)^t

v(t) = 173(0.955)^t

After 20 years :

v(t) = 173(0.955)^t

v(20) = 173(0.955)²⁰

= 68.88

Problem 9 :

A plague of mice has hit Australia! Starting with only 30 mice, their population 𝑝 increases by 650% every month, 𝑚.

Solution :

Exponential growth :

m(t) = p(1 + r%)^m

p = 30, r = 650 %

m(t) = 30(1 + 650%)^m

= 30(1 + 6.5)^m

= 30(7.5)^m

Problem 10 :

The rodent population 𝑝 in a large city is being controlled by a new poison that kills half the population every 6 months 𝑚. There are currently 2,000,000 rodents.

Solution :

A = A₀ (1/2) ^t/h

Here A₀ = 2,000,000, h = 6

A = 2,000,000 (1/2) ^t/6