The Exponential and Logarithmic Functions
[ad_1]Introduction
An important class of nonlinear functions that is of particular interest in economics
comprises the exponential and logarithmic functions. These functions
are useful for investigating problems associated with economic growth and decay
and mathematical problems in finance such as the compounding of interest
on an investment or the depreciation of an asset. For example, if a person invests
£3,000 in an investment bond for which there is a guaranteed annual rate
of interest of 5% for two years, the evaluation of an exponential function will
provide the return at the end of that period. If a credit card company charges
interest on an outstanding balance, the evaluation of an exponential function
will provide information on the AER (annual equivalent rate). We begin this
chapter by sketching the graphs of some exponential functions and highlighting
some of their important properties. Exponential functions are functions in
which a constant base a is raised to a variable exponent x. The general form
of an exponential function is given by
y = ax, where a > 0 and a = 1. (5.1)
The parameter a is known as the base of the exponential function. The independent
variable x occurs as the exponent of the base.
87
88 Elements of Mathematics for Economics and Finance
5.2 Exponential Functions
All exponential functions of the form f(x) = ax satisfy the following properties:
Properties
- The domain of f(x) is the set of all real numbers; the range of f(x) is the
set of all positive real numbers. - For all a > 1, f(x) is increasing; for 0 < a < 1, f(x) is decreasing.
- For all a > 0 with a = 1, f(0) = 1.
- For a > 1, f(x) tends to 0 as x tends to −∞; for 0 < a < 1, f(x) tends to
0 as x tends to +∞. - For a > 1, f(x) tends to +∞ (i.e., increases without bound) as x tends to
+∞; for 0 < a < 1, f(x) tends to +∞ as x tends to −∞. In Fig. 5.1, the graphs of y = 2x and y = 2−x = (1 2 )x are sketched for −4 ≤ x ≤ 4. These graphs illustrate some of the properties of exponential functions. Clearly, the domain of both functions is the entire real line, and the range is the set of all positive real numbers. The graph of f(x) = 2x is strictly increasing and f(x) tends to +∞ as x tends to +∞. The graph of f(x) = 2−x is strictly decreasing and f(x) tends to 0 as x tends to +∞. Note also that the graphs of y = 2x and y = 2−x are reflections of each other in the y-axis under the reflection (x, y) → (−x, y). In Fig. 5.2, the graphs of two exponential functions with bases a = 2 and a = 5 are sketched. This figure shows that, for bases a1, a2 satisfying a2 >
a1 > 1, ax2
increases in value faster than ax1
for x > 0.
An important base that is useful in many areas of mathematics as well as in
applications to problems in economics is the irrational number e, whose most
significant digits are given by
e = 2.7182818284 . . . .
This mathematical constant like the constant π does not have a finite decimal
representation and is another example of an irrational number. Its decimal form
is therefore never ending and is not a repeating decimal. It is interesting to see
how this number can be defined without going into the mathematical details.
Consider the function
f(x) =
1 +
1
x
x
. - The Exponential and Logarithmic Functions 89
x
-4 -2 0 2 4
5
10
15
y
Figure 5.1 The graph of the functions f(x) = 2x (continuous curve) and
f(x) = 2−x (dashed curve).
Let us evaluate this function for increasing values of x, for example x = 1, 10,
100, 1,000, and 10,000.
x f(x)
1
1 +
1
1
1
= 2
10
1 +
1
10
10
= 2.593742460
100
1 +
1
100
100
= 2.704813829
1,000
1 +
1
1,000
1000
= 2.716923932
10,000
1 +
1
10,000
10000
= 2.71815
These calculations show that as x gets larger, the value of
1 +
1
x
x
90 Elements of Mathematics for Economics and Finance
x
-4 -2 0 2 4
5
10
15
20
25
y 30
Figure 5.2 The graph of the functions f(x) = 2x (continuous curve) and
f(x) = 5x (dashed curve).
increases and approaches a limiting value of 2.718281828 . . ., which traditionally
is denoted by the letter e. Mathematically, we write
e = lim
x→∞
1 +
1
x
x
,
i.e., as x approaches infinity, the value of the function f(x) =
1 + 1
x
x approaches
the constant e. The graph of this function is plotted in Fig. 5.3 where
the dotted line corresponds to the straight line y = e. In this figure we see that,
as x increases, f(x) gradually approaches the dashed line.
5.3 Logarithmic Functions
Logarithms have inspired a feeling of dread in generations of students on their
first encounter with them. Logarithmic functions are closely related to exponential
functions and it is this relationship that we will exploit in our description of
some of their key properties. Logarithms are useful for simplifying calculations
involving economic functions. If we take the exponential function defined by - The Exponential and Logarithmic Functions 91
x
20 40 60 80 100
0
0.5
1
1.5
2
2.5
y 3
Figure 5.3 The graph of the function f(x) =
1 + 1
x
x. The dashed line
corresponds to the constant function f(x) = e.
y = ax and interchange the dependent variable y with the independent variable
x, we obtain
x = ay.
This defines a new function y = loga x, known as the logarithmic function
with base a, which is the exponent to which a must be raised to get x, i.e.,
x = ay ⇔ y = loga x.
Thus, the logarithmic function y = loga x is the inverse of the exponential
function y = ax. For example, if we wish to evaluate y = log10 100, then
100 = 10y. Since 100 = 102, we find that y = 2 so that log10 100 = 2. The
restrictions on the base are the same as for the exponential functions, i.e.,
a > 0, a = 1.
There are two important bases:
• a = 10 gives rise to common logarithms, written simply as log x.
• a = e where e ≈ 2.71828 gives rise to natural logarithms, written as ln x.
Common and natural logarithms may be evaluated numerically by pressing
either the log or ln keys, respectively, on a scientific calculator. For example, to
evaluate log 2.5:
92 Elements of Mathematics for Economics and Finance
x
2 4
-4
-2
0
2
4
y
Figure 5.4 The graph of the functions f(x) = log2 x (continuous curve)
and f(x) = log1/2 x (dashed curve). - Enter 2.5
- Press the log key
You should obtain the answer 0.397940009 to 9 decimal places, i.e., log 2.5 =
0.397940009. Note that on some calculators you press the log key first, then
enter the number and finally press the = key. Similarly, to evaluate the natural
logarithm of 2.5: - Enter 2.5
- Press the ln key
In this case, you should obtain the answer 0.916290732 to 9 decimal places, i.e.,
ln 2.5 = 0.916290732.
Properties of the function f(x) = loga x - The domain of the function is the set of all positive real numbers; the range
is the set of all real numbers. - For base a > 1, f(x) is increasing. For 0 < a < 1, f(x) is decreasing.
- The Exponential and Logarithmic Functions 93
Table 5.1 Table of values of log2 x and log1/2 x.
x 1/4 1/2 1 2 4
log2 x −2 −1 0 1 2
log1/2 x 2 1 0 −1 −2 - At x = 1, y = 0 independent of the base.
The graphs of the logarithmic functions y = log2 x and y = log1/2 x are
shown in Fig. 5.4. These logarithmic functions may be written equivalently as
x = 2y and x = (1/2)y, respectively, and are tabulated in Table 5.1. Note that
these graphs are reflections of the graphs of y = 2x and y = 2−x, respectively,
in the line y = x.
Example 5.1
Evaluate the following: - log8 64,
- log3( 1
81 ), - log16 2.
Solution. - Let y = log8 64, then 8y = 64 = 82 and so y = 2.
- Let y = log3
1
81 , then 3y = 1
81 = 1
34 = 3−4 and so y = −4. - Let y = log16 2, then 16y = 2 or (24)y = 24y = 2 and so 4y = 1 and
therefore y = 1
4 .
Example 5.2
Solve the following for x: - log4 x = 3,
- log81 x = 3
4 .
Solution. - x = 43 = 64.
94 Elements of Mathematics for Economics and Finance - x = 813/4 = (811/4)3 = 33 = 27.
Rules of Logarithms
For a, x, y positive real numbers, and n a real number, and base a = 1:
loga(xy) = loga x + loga y, (5.2)
loga(x/y) = loga x − loga y, (5.3)
loga xn = n loga x, (5.4)
loga
n √
x = loga
x1/n =
1
n
loga x. (5.5)
Note that loga x2 means the logarithm of x2 and not the square of loga x, which
is written as log2 x.
To prove the first two rules, let s = loga x and t = loga y. Using the relationship
between the logarithmic and exponential functions, we have x = as
and y = at. Then using the product rule for exponents, we obtain
xy = asat = as+t.
So s + t is the power to which the base must be raised to give xy, i.e.,
s + t = loga x + loga y = loga(xy).
Similarly, using the quotient rule for exponents, we have
x
y
= as
at = as−t.
So s − t is the power to which the base must be raised to give x/y, i.e.,
s − t = loga x − loga y = loga(x/y).
Example 5.3
Solve the equation ln(x + 4)2 = 3 for x.
Solution.
2 ln(x + 4) = 3
ln(x + 4) =
3
2
x + 4 = e1.5
x + 4 = 4.48169 to 5 decimal places
x = 0.48169 to 5 decimal places - The Exponential and Logarithmic Functions 95
Example 5.4
Express loga 3 + loga 4 − loga 6 as a single logarithm.
Solution.
loga 3 + loga 4 − loga 6 = loga(3 × 4) − loga 6
= loga
3 × 4
6
= loga
2
Example 5.5
Find the value of x satisfying
loga x = 3 loga 2 + loga 20 − loga 1.6.
Solution.
loga x = 3 loga 2 + loga 20 − loga 1.6
= loga
23 + loga 20 − loga 1.6
= loga
8 × 20
1.6
= loga
100
Therefore x = 100.
5.4 Returns to Scale of Production Functions
The output, Q, of any production process depends on a variety of inputs, known
as factors of production. These include land, capital, labour, and enterprise.
For simplicity, here we restrict our attention to capital, K, and labour, L. The
dependence of Q on K and L is indicated by writing
Q = Q(K,L).
Q is called a production function. It is an example of a function of two
variables – in this case K and L. Functions of two variables are described in
more detail in Chapter 8.
96 Elements of Mathematics for Economics and Finance
If Q(K,L) = 100K1/3L1/2, then when K = 27 and L = 100 the output
Q(27, 100) is given by
Q = 100(27)1/3(100)1/2
= 100(3)(10)
= 3,000
Of particular interest is what happens to the output when the inputs are scaled
in some way. If capital and labour double, does the production level double,
does it go up by more than double, or does it go up by less than double? For
the above production function, we see that when K and L are replaced by 2K
and 2L, respectively, then using the rules of indices (see Section 1.6):
Q = 100(2K)1/3(2L)1/2
= 100(21/3K1/3)(21/2L1/2)
= (21/321/2)(100K1/3L1/2)
= 25/6(100K1/3L1/2)
The term in brackets is just the original output. So this is multiplied by 25/6 ≈
1.78 so output goes up by just less than double when capital and labour are
doubled.
In general, a function
Q = Q(K,L)
is said to be homogeneous if
Q(λK, λL) = λnQ(K,L), (5.6)
for some number n where λ is a general number. The power, n, is called the
degree of homogeneity. Let us take the previous example again:
Q(λK, λL) = 100(λK)1/3(λL)1/2
= (λ1/3λ1/2)100K1/3L1/2
= λ5/6Q(K,L)
This production function is homogeneous of degree 5/6.
In general, if the degree of homogeneity, n, satisfies - n < 1 the function is said to display decreasing returns to scale.
- n = 1 the function is said to display constant returns to scale.
- n > 1 the function is said to display increasing returns to scale.
- The Exponential and Logarithmic Functions 97
5.4.1 Cobb-Douglas Production Functions
Functions of the form
Q = AKαLβ
where A, α, β are constants are called Cobb-Douglas production functions.
These are homogeneous of degree α + β since if
Q(K,L) = AKαLβ
then
Q(λK, λL) = A(λK)α(λL)β
= λα+β(AKαLβ)
= λα+βQ(K,L)
Therefore, Cobb-Douglas production functions exhibit - decreasing returns to scale if α +β < 1.
- constant returns to scale if α + β = 1.
- increasing returns to scale if α +β > 1.
Example 5.6
Show that the production function
Q = K2 + 3KL,
is homogeneous and comment on its returns to scale.
Solution. In this example we are given that
Q = Q(K,L) = K2 + 3KL.
If we scale or multiply both K and L by λ, then the corresponding value of
output is
Q(λK, λL) = (λK)2 + 3(λK)(λL)
= λ2K2 + 3λ2KL
= λ2(K2 + 3KL)
= λ2Q(K,L).
Therefore, we have shown that
Q(λK, λL) = λ2Q(K,L),
98 Elements of Mathematics for Economics and Finance
which on comparison with (5.6) demonstrates that the production function is
homogeneous of degree 2. Since the degree of homogeneity is greater than one,
the function displays increasing returns to scale.
The Cobb-Douglas production function is an example of a nonlinear function.
However, it may be converted to a linear function through a simple logarithmic
transformation as follows. Take natural logarithms of both sides of the
equation
Q = AKαLβ.
Then
lnQ = ln(AKαLβ)
= lnA + lnKα + lnLβ
= lnA + α lnK + β ln L.
This is what we call a log-linear function. If we define ˜Q = lnQ, ˜K = lnK,
and ˜L = lnL, then
˜Q
= lnA + α˜K + β ˜L,
a linear function in the variables ˜K and ˜L.
5.5 Compounding of Interest
There is a plethora of investment products and loan facilities available to an
individual in the financial market place. It is important for both an individual
or a business to make an informed choice between the financial products on
offer in order to maximize the return on their capital or to minimize the interest
on their loan repayments, for example. Suppose that a person wants to borrow
some capital and is offered two loan products. The first charges interest on the
loan at the annual rate of 12% while the second charges interest at a monthly
rate of 1%. Which product should the person go for? In this section, we show
how such decisions can be made.
Suppose that an individual wishes to invest a sum of £10,000 over a period
of three years and that the annual rate of interest is 5%. After one year, the
interest on the investment amounts to 5% of £10,000, which is £500. If the
investment is subject to simple interest, then the return on the investment
would be £500 per year for each subsequent year. The total amount of interest
earned over the five-year period in this case is 5 × £500 = £2,500. However,
most financial investment products use compound interest as a means of
enticing their customers not to withdraw the interest earned after the first and
subsequent years from the accumulated value of their investment.When interest - The Exponential and Logarithmic Functions 99
is compounded annually, the amount of interest earned in the second year is 5%
of £10,500, which is the sum of the initial investment (£10,000) and the first
year’s interest (£500). The interest earned in the second year is therefore £525
and so the value of the investment at the end of the second year is £10,500 +
£525 = £11,025. Finally, at the end of the third year the investment is worth
£11,025 plus 5% of £11,025 interest giving a total of £11,571.25.
There is a formula that can be used to determine the future value of an
investment. Let P0 denote the value of the initial investment. This is sometimes
known as the principal. Let Pt denote the value of the investment after t years.
If the interest on the principal is compounded annually, at an interest rate r
(written as a decimal or fraction), then after one year the investment is worth
P1 = P0 + rP0 = P0(1 + r). (5.7)
Similarly, after the second year the investment is worth
P2 = P1 + rP1 = P1(1 + r). (5.8)
Substituting for P1 in (5.8) using (11.1) we have
P2 = [P0(1 + r)](1 + r) = P0(1 + r)2. (5.9)
In general, one can show that
Pt = P0(1 + r)t. (5.10)
Now suppose that the interest is compounded semi-annually (six monthly
intervals). In this case, (5.10) would have to be modified to
Pt = P0
1 + r
2
2t
. (5.11)
Note that the differences between this formula and the formula (5.10) when
the interest is compounded annually are that the interest rate is divided by 2
since the interest is added twice a year and t is replaced by 2t since this is the
number of times that the interest is added during t years. Similarly, one can
show that if interest is added monthly, the value of the investment after t years
is
Pt = P0
1 + r
12
12t
. (5.12)
If this argument is continued and interest is compounded n times a year,
then we have the formula
Pt = P0
1 + r
n
nt
. (5.13)
100 Elements of Mathematics for Economics and Finance
If n is very large then we are approaching the situation in which interest is
added continuously (at every instant of time) instead of at discrete moments
in time. If we make the substitution m = n/r in (5.13), then we have
Pt = P0
1 +
1
m
mrt
= P0
1 +
1
m
m
rt
(5.14)
We saw in Section 5.2 that
1 +
1
m
m
→ e as m→∞.
If we allow m → ∞ in (5.14) (which is equivalent to allowing n → ∞ in
(5.13) since r is held constant), then we obtain the formula for the continuous
compounding of interest:
P(t) = P0ert. (5.15)
In this formula, t need no longer be a positive integer. It can take any positive
value.
For negative growth rates, such as depreciation or deflation, the same formulae
apply but with t or r negative.
Example 5.7
Suppose that the sum of e100 is invested at an annual rate of interest of
10%. Calculate the value of the investment in five years’ time if the interest is
compounded (a) annually, (b) semi-annually, (c) continuously.
Solution. - We apply the formula (5.10) with P0 = 100, r = 10% = 0.1 and t = 5.
Inserting these values into the formula gives
P5 = 100(1 + 0.10)5 = e161.05. - We apply the formula (5.11) with P0 = 100, r = 10% = 0.1 and t = 5.
Inserting these values into the formula gives
P5 = 100
1 +
0.10
2
2×5
= 100(1.05)10 = e162.89. - The Exponential and Logarithmic Functions 101
- We apply the formula (5.15) with P0 = 100, r = 10% = 0.1 and t = 5.
Inserting these values into the formula gives
S = 100e0.10×5 = 100e0.2 = e164.87.
Example 5.8
The value of an asset, currently priced at $250,000, is expected to increase by
12% a year. - Find its value in ten years’ time.
- After how many years will it be worth at least 1.25 million dollars?
Solution. - We use the formula (5.15) with P0 = 250, 000, r = 12% = 0.12, and t = 10.
Inserting these values into the formula yields
P10 = 250,000(1 + 0.12)10
= 250,000(1.12)10
= $776,462.05
Therefore, after 10 years the asset will be worth $776,462.05. - In this part of the question, we use the formula (5.10) again but this time
we know Pt = 1,250,000 and we need to determine the value of t. We need
to find the value of t for which
1,250,000 = 250,000(1 + 0.12)t
5 = (1.12)t
Take natural logarithms of both sides:
ln 5 = ln(1.12)t = t ln 1.12.
Therefore,
t = ln5/ ln 1.12 = 14.20
So after 15 years, the asset will be worth at least 1.25 million dollars.
Example 5.9
A credit card company charges interest at 2% per month. What is the annual
equivalent rate correct to two decimal places?
102 Elements of Mathematics for Economics and Finance
Solution. Suppose that the balance outstanding on the credit card is B then
the amount owing (loan plus interest) over one year is
B
1 +
2
100
12
= B(1.02)12.
Here we have used the formula (5.12) but in which we have not divided r by
12 since the rate of interest is already a monthly one. Let R be the annual
equivalent rate. Then if interest is charged annually, the amount owing after a
year is
B
1 + R
100
.
Equating these two expressions enables us to find R:
B(1.02)12 = B
1 + R
100
.
Therefore,
(1.02)12 = 1+ R
100.
Rearranging this equation gives
R = [(1.02)12 − 1] × 100 = 0.2682 = 26.82%.
This is the annual equivalent rate.
5.6 Applications of the Exponential Function in
Economic Modelling
Example 5.10
During a recession, a firm’s revenue declines continuously at an annual rate of
10% so that total revenue (measured in millions of pounds) in t years’ time is
modelled by
TR = 8e
−0.1t. - Calculate the current revenue and also the revenue in two years’ time.
- Sketch the graph of TR against t.
- Rearrange the formula to get t in terms of TR.
- After how many years will revenue decline to below £5 million?
- The Exponential and Logarithmic Functions 103
Solution. - When t = 0, TR = 8e0 = 8. Therefore, the current revenue is £8 million.
When t = 2, TR = 8e−(0.1)(2) = 8e−0.2 = 6.55. Therefore, after two years
the revenue will have declined to £6.55 million. - The graph of the revenue function TR plotted as a function of time t is
shown in Fig. 5.5. - The first step in the process of rearranging the formula for TR is to divide
both sides by 8:
TR
8
= e
−0.1t.
Then taking natural logarithms of both sides:
ln
TR
8
= lne
−0.1t = −0.1t.
Finally, dividing both sides by −0.1, we obtain the formula for t in terms
of TR:
t =
1
(−0.1)
ln
TR
8
= −10 ln
TR
8
. - We now use this formula to determine the number of years after which the
revenue will decline to £5 million. Inserting TR = 5 in this formula yields
t = −10 ln
5
8
= 4.700 (to 3 decimal places).
Therefore, after 5 years TR will decline to £5 million. An estimate for this
answer can be found from the graph in Fig. 5.5. The dashed line in this
graph corresponds to TR = 5. The intersection of this straight line with
the curve TR = 8e−0.1t provides the answer.
Example 5.11
The percentage, y, of Europeans possessing a mobile phone t years after it was
introduced is modelled by
y = 80 − 70e
−0.2t. - Find the percentage of Europeans that have mobile phones
a) at the launch of the product;
b) after 3 years;
c) after 10 years.
104 Elements of Mathematics for Economics and Finance
t
5 10 15 20
0
2
4
6
8
10
TR
Figure 5.5 The graph of the function TR = 8e−0.1t. The dashed line
corresponds to TR = 5. - What is the market saturation level?
- After how many years will the percentage of Europeans possessing mobile
phones first reach 75%?
Solution. - a) The launch of the product corresponds to t = 0 since t measures the
time from the introduction of mobile phones into the market place. So
putting t = 0 into the expression for y gives
y = 80 − 70e0 = 80 − 70 = 10%.
b) After three years t = 3, the percentage of Europeans possessing mobile
phones is given by
y = 80 − 70e
−0.2×3 = 80 − 70e
−0.6 = 41.58%.
c) After ten years t = 10, the percentage of Europeans possessing mobile
phones is given by
y = 80 − 70e
−0.2×10 = 80 − 70e
−2 = 70.53%. - The Exponential and Logarithmic Functions 105
t
5 10 15 20
0
20
40
60
80
y
Figure 5.6 The graph of the function y = 80 − 70e−0.2t. - The market saturation is the limiting value of y as t tends to ∞. Since
e
−0.2t → 0 as t→∞,
we have
y → 80 as t→∞,
and so the market saturation level is 80% (see Fig. 5.6). - To determine the time after which 75% of Europeans possess a mobile
phone, we rearrange the equation and express t in terms of y and then put
y = 75 into the resulting formula. A simple rearrangement gives
e
−0.2t =
(80 − y)
70 .
Taking natural logarithms of both sides yields
−0.2t = ln
80 − y
70
.
106 Elements of Mathematics for Economics and Finance
Finally, multiplying both sides by 1/(−0.2) = −5 gives
t = −5 ln
80 − y
70
= −5 ln
5
70
= 13.20
Therefore, after 14 years the percentage of Europeans possessing mobile
phones will break through the 75% barrier.
EXERCISES
5.1. Sketch the functions y = 2x and y = 3x on the same graph for
−2 ≤ x ≤ 2.
5.2. Evaluate (a) log3 9, (b) log4 2, (c) log7(1/7).
5.3. Show that the following production functions are homogeneous and
comment on their returns to scale:
a) Q = 7KL2,
b) Q = 50K1/4L3/4.
5.4. Determine the annual rate of interest required for a principal of
£2,000 to produce a value of £10,000 after 8 years.
5.5. Determine the annual equivalent rate (AER) corresponding to a
monthly rate of 1.15%.
5.6. An economy is forecast to grow continuously at an annual rate of
3% so that the gross national product (GNP), measured in billions
of euros, after t years is given by
GNP = 60e0.03t.
a) Calculate the current value of GNP and its future value in four
years’ time.
b) After how many years is GNP forecast to be 90 billion euros?
5.7. Determine the rate of interest required for an investment that is currently
worth $1,000 to be worth $4,000 after 10 years if the interest
is compounded continuously. - The Exponential and Logarithmic Functions 107
5.8. Determine the annual equivalent rate (AER) corresponding to a
monthly rate of 1%.
5.9. The percentage, y, of households possessing dishwashers t years after
they have been introduced in a country is modelled by
y = 30 − 25e
−0.2t.
a) Find the percentage of households that have dishwashers
i. at their launch;
ii. after 1 year;
iii. after 10 years;
iv. after 20 years.
b) What is the market saturation level?
c) After how many years will the percentage of households possessing
dishwashers first reach 15%?
5.10. A firm’s turnover, y, measured in millions of pounds, after t years is
given by
y = 8e0.09t.
What is its turnover in its initial year of trading and after two years
of trading? After how many years will its turnover have doubled
since it started trading?
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