Representing Real-Life Situations Using
Expon
by CHED on June 15, 2017
lesson duration of 5 minutes
under General Mathematics
generated on June 15, 2017 at 12:07 pm
Tags: Exponential Functions, Sunlight, Trees, Plants
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Generated: Jun 15,2017 08:07 PM
Representing Real-Life Situations Using Expon
( 5 mins )
Written By: CHED on May 20, 2016
Subjects: General Mathematics
Tags: Exponential Functions, Sunlight, Trees, Plants
Resources
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Content Standard
The learner demonstrates understanding of key concepts of inverse functions, exponential functions, and logarithmic
functions.
Performance Standard
The learner is able to apply the concepts of inverse functions, exponential functions, and logarithmic functions to
formulate and solve real-life problems with precision and accuracy.
Learning Competencies
The learner solves problems involving exponential functions, equations, and inequalities.
The learner represents real-life situations using exponential functions.
The learner solves problems involving exponential functions, equations, and inequalities.
Motivation 1 mins
Exponential functions occur in various real world situations. Exponential functions are used to model real-life situations
such as population growth, radioactive decay, carbon dating, growth of an epidemic, loan interest rates, and
investments.
Group Activity. This activity can help introduce the concept of an exponential function.
Materials. One 2-meter string and a pair of scissors for each group
(a) At Step 0, there is 1 string.
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(b) At Step 1, fold the string into two equal parts and then cut at the middle. How many strings of equal length do you
have? Enter your answer in the table below.
(c) At Step 2, again fold each of the strings equally and then cut. How many strings of equal length do you have? Enter
your answer in the table below.
(d) Continue the process until the table is completely filled-up.
Questions.
(a) What pattern can be observed from the data?
(b) Define a formula for the number of strings as a function of the step number.
Answers.
It can be observed that as the step number increases by one, the number of strings doubles. If n is
the number of strings and s is the step number, then n = 2s.
Lesson Proper 1 mins
Definition.
An exponential function with base b is a function of the form f(x) = bx or y = bx,
bx, where b > 0 , b ?= 1.
EXAMPLE 1. Complete a table of values for x = 3, 2, 1, 0, 1, 2 and 3 for the exponential functions
Solution. The solution is as follows:
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EXAMPLE 2.
Solution.
Since ? ? 3.14159 is irrational, the rules for rational exponents are not applicable. We define 3? using rational
numbers: f(?) = 3? can be approximated by 33.14.
33.14. A better approximation is 33.14159.
33.14159. Intuitively, one can obtain any
level of accuracy for 3? by considering sufficiently more decimal places of ?. Mathematically, it can be proved that
these approximations approach a unique value, which we define to be 3?.
3?.
Definition.
Let b be a positive number not equal to 1. A transformation of an exponential function with base b is a
function of the form
where a, c, and d are real numbers.
Many applications involve transformations of exponential functions. Some of the most common applications in real-life
of exponential functions and their transformations are population growth,
growth, exponential decay,
decay, and compound
interest.
(a) Populations
On several instances, scientists will start with a certain number of bacteria or animals and watch how the population
grows. For example, if the population doubles every 3 days, this can be represented as an exponential function.
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EXAMPLE 3. Let t = time in days. At t = 0, there were initially 20 bacteria. Suppose that the bacteria doubles every
100 hours. Give an exponential model for the bacteria as a function of t.
Solution. An exponential model for this situation is
Exponential Models and Population Growth.
Suppose a quantity y doubles every T units of time. If y0 is the initial amount, then the quantity y after t units
of time is given by
(b) Exponential Decay
Exponential functions can be used to model radioactive decay.
Definition.
The half-life of a radioactive substance is the time it takes for half of the substance to decay.
EXAMPLE 4. Suppose that the half-life of a certain radioactive substance is 10 days and there are 10g initially,
determine the amount of substance remaining after 30 days.
Solution. Let t = time in days. We use the fact that the mass is halved every 10 days (from definition of half-life). Thus,
we have:
An exponential model for this situation is
(c) Compound Interest
A starting amount of money (called the principal)
principal) can be invested at a certain interest rate that is earned at the end of
a given period of time (such as one year). If the interest rate is compounded,
compounded, the interest earned at the end of the
period is added to the principal, and this new amount will earn interest in the next period. The same process is
repeated for each succeeding period: interest previously earned will also earn interest in the next period. Compound
interest will be discussed in more depth in Lessons 23 and 25.
EXAMPLE 5. Mrs. De la Cruz invested P100,000. in a company that offers 6% interest compounded annually. How
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much will this investment be worth at the end of each year for the next five years?
Solution.
Solution. Let t be the time in years. Then we have:
An exponential model for this situation is
Compound Interest.
If a principal P is invested at an annual rate of r, compounded annually, then the amount after t years is given
by
EXAMPLE 6. Referring to Example 5, is it possible for Mrs. De la Cruz to double her money in 8 years? in 10 years?
Solution. Using the model y = 100000(1.06)t, substitute t = 8 and t = 10:
If t = 8, then y = P100,000(1.06)8 ? P159,384.81. If t = 10, then y = P100,000(1.06)10 ? P179, 084.77. Since her
money still has not reached P200, 000 after 10 years, then she has not doubled her money during this time.
The Natural Exponential Function
While an exponential function may have various bases, a frequently used based is the irrational number e ? 2.71828.
The enrichment in Lesson 27 will show how the number e arises from the concept of compound interest. Because e is
a commonly used based, the natural exponential function is defined having e as the base.
Definition.
The natural exponential function is the function f(x) = ex.
EXAMPLE 7. A large slab of meat is taken from the refrigerator and placed in a pre-heated oven. The temperature T of
the slab t minutes after being placed in the oven is given by
degrees Celsius. Construct a table of
values for the following values of t: 0, 10, 20, 30, 40, 50, 60, and interpret your results. Round off values to the nearest
integer.
Solution. The solution is as follows:
The slab of meat is increasing in temperature at roughly the same rate.
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Seatwork 1 mins
Seatwork 1. Suppose that a couple invested P50,000 in an account when their child was born, to prepare for the
child's college education. If the average interest rate is 4.4% compounded annually, (a) give an exponential model for
the situation, and (b) will the money be doubled by the time the child turns 18 years old?
Solution. An exponential model for the situation is A = 50, 000(1.044)t, where t is the number of years since the
amount was invested. If t = 18, then A = 50, 000(1.044)18 ? P108, 537.29. By the time the child turns 18 years old, the
money has more than doubled since the amount exceeded P100, 000.
Seatwork 2. You take out a P20,000 loan at a 5% interest rate. If the interest is compounded annually, (a) give an
exponential model for the situation, and (b) how much will you owe after 10 years?
Solution. An exponential model for this situation is A = P20,000(1.05)t where t is the number of years since the
amount was loaned. If t = 10, A = P20, 000(1.05)10 ? P32, 577.89. Thus, the total amount you owe after 10 years is
P32, 577.89.
Seatwork 3. Suppose that the half-life of a substance is 250 years. If there were initially 100 g of the substance, (a)
give an exponential model for the situation, and (b) how much will remain after 500 years?
Solution. The situation can be modeled by
where x = 0 corresponds to the time when there were
100 g of substance. If x = 500, then y =
= 25. Thus, there 22 will be 25 g remaining after
500 years.
Alternate Solution. After 250 years, the substance reduces to half (50 g); after another 250 years, the substance
reduces to half again (25 g).
Seatwork 1 mins
Seatwork 1. Suppose that a couple invested P50,000 in an account when their child was born, to prepare for the
child's college education. If the average interest rate is 4.4% compounded annually, (a) give an exponential model for
the situation, and (b) will the money be doubled by the time the child turns 18 years old?
Solution. An exponential model for the situation is A = 50, 000(1.044)t, where t is the number of years since the
amount was invested. If t = 18, then A = 50, 000(1.044)18 ? P108, 537.29. By the time the child turns 18 years old, the
money has more than doubled since the amount exceeded P100, 000.
Seatwork 2. You take out a P20,000 loan at a 5% interest rate. If the interest is compounded annually, (a) give an
exponential model for the situation, and (b) how much will you owe after 10 years?
Solution. An exponential model for this situation is A = P20,000(1.05)t where t is the number of years since the
amount was loaned. If t = 10, A = P20, 000(1.05)10 ? P32, 577.89. Thus, the total amount you owe after 10 years is
P32, 577.89.
Seatwork 3. Suppose that the half-life of a substance is 250 years. If there were initially 100 g of the substance, (a)
give an exponential model for the situation, and (b) how much will remain after 500 years?
Solution. The situation can be modeled by
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where x = 0 corresponds to the time when there were
100 g of substance. If x = 500, then y =
= 25. Thus, there 22 will be 25 g remaining after
500 years.
Alternate Solution. After 250 years, the substance reduces to half (50 g); after another 250 years, the substance
reduces to half again (25 g).
Evaluation 1 mins
Solve the given problems and check.
(a) A population starts with 1,000 individuals and triples every 80 years. (a) Give an exponential model for the
situation. (b) What is the size of the population after 100 years?
Solution. The situation can be modeled by
where x = 0 corresponds to the time when there were
1,000 individuals. If x = 100, then y =
Thus, there will be 3,948 individuals after
100 years.
(b) P10, 000 is invested at 2% compounded annually. (a) Give an exponential model for the situation. (b) What is the
amount after 12 years?
Solution. The situation can be modeled by A = 10,000(1.02)t, where t is the number of years since the amount was
invested.
Thus, the amount after 12 years is P12, 682.42.
Download Teaching Guide Book 0 mins
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