WORD PROBLEMS ON EXPONENTIAL FUNCTIONS

Problem 1 :

A cup of green tea contains 35 milligrams of caffeine. The average teen can eliminate approximately 12.5% of the caffeine from their system per hour.

a) Write an exponential function to represent the amount of caffeine remaining after drinking a cup of green tea.

b) Estimate the amount of caffeine in a teenager’s body 3 hours after drinking a cup of green tea.

Solution :

a)  a = 35

Per hour 12.5 % of caffeine is being eliminated each hour.

then, r = 12.5

y = 35(1 - 12.5%)^x 

Here x is the number of hours.

y = 35(1 - 0.125)^x 

y = 35(0.875)^x 

b) Number of hours = 3

y = 35(0.875)³ 

= 35(0.6699) 

y = 23.45

So, the amount caffeine after 3 hours is 23.45 milligrams

Problem 2 :

From 1990 to 2000, the population of California can be modeled by

P = 29,816,591(1.0128)^t

where t is the number of years since 1990. Estimate the population in 2007.

Solution :

To estimate the population in 2007, we have to apply t as 17.

P = 29,816,591(1.0128)¹⁷

= 37013551.9

Approximately 37013552.

So, the required population at 2007 is 37013552..

Problem 3 :

You buy a new car for $22,500. The value of the car decreases by 25% each year.

Write an exponential decay model giving the car's value V (in dollars) after t years. What is the value of the car after three years?

Solution :

Initial value = 22500

Decreasing rate = 25%

y = 22500(1 - 25%)^x 

= 22500(1- 0.25)^x 

= 22500(0.75)^x 

Problem 4 :

A virus spreads through a network of computers such that each minute, 25% more computers are infected. If the virus began at only one computer, find the model for this situation and find the number of computers affected after 40 minutes.

Solution :

Number of virus in the initial stage = 1

Increasing factor = 25%

y = 1(1 + 25%)^x 

= 1(1 + 0.25)^x 

= 1(1.25)^x 

The total number of computer after 40 minutes.

= 1(1.25)⁴⁰

= 7523

Problem 5 :

A radioactive substance decays exponentially. A scientist begins with 100 milligrams of a radioactive substance. After 35 hours, 50 mg of the substance remains. How many milligrams will remain after 54 hours?

Solution :

Exponential function for growth or decay

A(t) = ae^rt ---(1)

Here A(t) is the quantity of substance after t hours.

Initial value = 100

100 = a(e)^r(0) 

100 = a(e)⁰

100 = a(1)

a = 100

Applying the value of a in (1), we get

A(t) = 100e^rt

Here t = 35, A(t) = 50

50 = 100e^r(35)

1/2 = e^r(35)

ln (1/2) = 35r

r = ln(0.5)/35

r = -0.693/35

r = -0.019

Approximately r = -0.02

Rate of decreasing = 2%

Quantity of substance after 54 hours.

A(t) = 100e^{(-0.02) 54}

= 100e^(-1.08)

= 33.96

Approximately 34 milligrams.

Problem 6 :

A radioactive substance decays exponentially. A scientist begins with 110 milligrams of a radioactive substance. After 31 hours, 55 mg of the substance remains. How many milligrams will remain after 42 hours?

Solution :

Exponential function for growth or decay

A(t) = ae^rt ---(1)

Here A(t) is the quantity of substance after t hours.

Initial value = 110

110 = a(e)^r(0) 

110 = a(e)⁰

110 = a(1)

a = 110

Applying the value of a in (1), we get

A(t) = 110e^rt

Here t = 31, A(t) = 55

55 = 110e^r(31)

1/2 = e^r(31)

ln (1/2) = 31r

r = ln(0.5)/31

r = -0.693/31

r = -0.022

Approximately r = -0.022

Rate of decreasing = 2.2%

Quantity of substance after 42 hours.

A(t) = 110e^{(-0.022) 42}

= 110e^(-0.924)

= 110/2.519

= 43.66

Approximately 44 milligrams.

Problem 7 :

A house was valued at $110,000 in the year 1985. The value appreciated to $145,000 by the year 2005. What was the annual growth rate between 1985 and 2005?

Assume that the house value continues to grow by the same percentage. What did the value equal in the year 2010?

Solution :

By observing the given situation, the required function will be exponential growth function.

P(t) = a(1 + r%)^x 

At 1985, the population is $110,000

110000 = a(1 + r%)⁰ 

a = 110000

Applying the initial value in (1), we get

P(t) = 110000(1 + r%)^x 

In the year 2005, the value of x = 20

145000 = 110000(1 + r%)²⁰ 

(1 + r%)²⁰  = 145000/110000

(1 + r%)²⁰ = 1.318

1 + r% = (1.318)^1/20

1 + r% = (1.318)^0.05

r% = 1.014 - 1

r% = 0.014

r = 1.4

In the year 2010, the value of x = 25

P(t) = 110000(1 + 1.4%)²⁵ 

= 110000(1.416)

= 155760

So, the population at 2010 is 155760.