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12. EXPONENTIAL AND LOGARITHMIC FUNCTION
Exponential and logarithmic functions have many important applications in engineering,
biology, chemistry, geology, and many other fields. In this section we will define
exponential and logarithmic functions and study their properties. In particular, we shall
see that these functions are inverses of each other.
Exponential function
Suppose that b is any fixed positive real number, and suppose R is the set of real
numbers.
DEFNITION
if b > 0 and x e R is any real number, then the function defined by
is called the EXPONENTIAL FUNCTION with base b .
The DOMABI of f (x) = b x is the set of all real numbers.
The RANGE of the exponential function is the set of all POSITIVE real numbers .
We know the laws of exponents for rational exponents. ( see Chapter 5 : Exponents and
their properties.)
We recall that if b > 0 and c and d are arbifrary rational numbers, then
b c .b d —b C + d — = bC-d
b 0 =1
1
b—c
With the definition above the Laws ofExponents hold for all REAL EXPONENTS.
( It will be proved next year .)
110
EXAMPLE 1
Sketch the graph of the exponential function
SOLUTION : Let us make a brief table of values :
x101112131 I-h 1-1/3 11Y 4181 h 11/81
9
8
7
6
5
4
3
2
1
Now we list several important facts regarding this function and its graph.
l.) As x becomes very large and positive, f (x) = 2 x also becomes very large andpositive. We say that f (x) grows without bound as x grows without bound.
2.) As x grows numerically large in the negative direction, f (x) = 2 x approaches 0.
3.) Since 2 | , the graph crosses the y axis at the point (0 , 1).
4.) f (x) = 2 x is strictly increasing on its entire domain, making it a one-to-one
function.
This graph is typical ofexponential functions WITH BASE GREATER THAN 1 .
1.) Since 1 , each ofthese graphs has (0 , 1) as its Y-INTERCEPT .
2.) For each of these graphs, as x —+ + , b x rapidly becomes extremely large :
f(x) = b x — +Ø . This behavior is known as EXPONENTIAL GROWTH .
3.) Finally, for each of these graphs, as x , f (x) = b x--»0 (with b > I
That is, the x axis is a HORIZONTAL ASYVTTOTE ofthese graphs.
111
EXAMPLE 2
Sketch the graph of the exponential function
(1/2)x.
SOLUTION : Comparing the function
with the function f (x) 2 x , we note that the image of x under f (x) = (1/2) x will beexactly the same as the image of ( - x ) under the function 2 x .
Hence, the graph ofthese THE Y AXIS.
are REFLECTIONS OF EACH OTHER ACROSS
9
8
6
5
4
3
1
1 2 3 4
Now we list the properties of the graph :
1.) As x becomes very large and positive, f (x) = (1/2) x becomes very close tozero.
2.) As x becomes very large in the negative direction, f (x) becomes very largeand positive.
3.) The y intercept point is (0 , l) .
4.) f (x) = (1/2) x is strictly decreasing on its entire domain, making it a one-to-onefunction.
This graph is typical ofexponential functions WITH BASE LESS THAN 1
112
NOTICE the PROPERTIES ofthe EXPONENTLAL FUNCTIONS
a.) If b > 1 , f (x) = b x is a strictly increasing function.
b.) If 0 < b < 1 , f (x) = b x is a strictly decreasing function.
c.) All exponential functions f (x) = b x , regardless of their base, have a
common point of intersection : P ( 0, 1 ) .Because
if x = 0 , then b 1 , for all values of b greater than zero.
EXERCISES
1. Sketch the graph of the functions :
a.) y=3X
c.) f (x) = (1/4) x
2. Tell which of the following statements are frue and which are false. Justify your
answers.
a.) b X - Y =b X -b Y b.) f (x) = b x is strictly decreasing
c.) f (x) = x 2 / 3 is an exponential function
d.) If c is a positive real constant and - oo < x < + , then f (x) = c x is strictly
increasing.
e.) The graph of f (x) = b x is symmetric about the y axis.
f.) The graph of f (x) = b x has no x— intercept point.
3. Determine which function is odd, even, or neither even nor odd :
d.) f (x) = x 2 +2 X
e.) f (x) =x.3 x
113
Logarithmic function
Since the exponential function is a one-to-one function, this function has an inverse.
DEFNITION :
If b > 0 and b 1 , then the function defined by
is called the LOGARITHMIC FUNCTION to the base b.
The logarithmic function to the base b is the inverse function of the exponential function
with base b .
That means,
y = log b x if and only if b Y =x.
This equality says that logarithms are exponents. Indeed
log b x is simply that exponent to which we should raise the base b in
order to obtain the positive number x .
For instance, log 2 8 is the exponent that the base 2 must be raised in order to obtain 8 ;that is ,
log 2 8 = 3 , since 2 3
Two important identities :
1.) y = log b x
2.) y=bX
if and only if
if and only if
b (log x)
log b (b X ) = x
EXERCISES
1. Put each equation in logarithmic form :
a.) 2 5 ̄ 32 .
c) 4 - = 1/64 d.)
2. Find each of the given logarithms :
a.) log 10 10 b.)
c.) log 0.0001 d.)
3. Find each of the given logarithms :
a.) log 82
c.) log 4 ( 1/2)
b.)
d.)
7 0 =1
= 81
log 100000
log 64 4
log 4 64
log 4 2
114
e.) log 16 16
g.) log 3 ( 1/81)
i.) log 8 32
k.) log 2 ( 1/16)
f.) log 5 25
h.) log 1/3 27
j.) log ( 1/27)
l.) log 4 1
4. Find the unknown N , b , u in each equation :
a.) log b 32=5
c.) log9N= l/2
e.) log 8 = u
g.) log b 1=0
i.) log 4N=-3
k.) log 5N=-2
m.) log b ( 1/7)=- 1/3
o.) log 4 (log 4 16) = u
r.) log 2 ( log
t.) 2 (log 3)
b.) log b 81 = 4
d.) log 8 N = 2/3
f.) log 32 8 = u
h.) log 55
j.) log b ( 1/9)
l.) log 3 81 = u
n.)log 681 = u
p.) log b ( log b 27 3 )
s.) log 4 N = 4
Since the logarithmic function is the inverse function of the exponential function ,
the DOMABI of the logarithmic function is ( 0 , + oo ) and
the RANGE of the logarithmic ftnction is ( - 00, + 00 )
Since they are inverse functions, the graph of the logarithmic function to the base b is thereflection across the line y = x of the exponential function with base b.
If b > 1 , y = b x is a strictly increasing function, hence
y = log b x is also a strictly increasing, function.
If O < b < 1 , y = b x is a strictly decreasing function, hence
y = log b x is also a strictly decreasing function.
115
EXAMPLE 3
Sketch the graph of
y = log 2 x and y = 2 x
(0,1)
on the same coordinate axes.
y=28
y=10g2x
(1,0)
Notice the properties of the logarithmic function :
a.) The x intercept point is ( 1 , 0) .
b.) The graph has no y intercept point, but the y axis is an asymptote of
the graph of y = log b x .
c.) All logarithmic functions f (x) = log b x , regardless of their base, havea common point of intersection. If x = 1 , then log b x = 0 , for all values of
b greater than zero.
116
EXAMPLE 4
Sketch on the same coordinate axes
y = log % x and y = (1/2) x
(0,1
Flog 1/2 X
EXERCISES
Sketch the graph of the functions
a.) y = 3 x and y = log 3 x
and y = log 1/5 (x)
117
Laws of logarithms
By using the definition of logarithms and the laws of exponents, three properties of
logarithms will be proved •
LAW 1.
The logarithm of the product of two positive numbers is equal to the sum of
the logarithms of the two numbers.
log b a.c = log b a + log b c
LAW 2.
The logarithm of the quotient of two positive numbers is equal to the
difference of logarithms of the two numbers.
alog b — = log b a— log b c
LAW 3.
The logarithm of a power of a positive number is equal to the exponent
times the logarithm of the number.
log b ac = c.log b a
PROOF: To prove LAW 1., let
x = log b a and y = log b c.
But by definition these two equations can be written in exponential form :
a = b X and c = b Y , respectively. Then
While changing this to logarithmic form yields
log b a.c = X + y.
Since x = log b a and y = log b c , the equation above becomes
log b a.c = log b a + log b c,
which is the desired result.
Laws 2. and 3. are proved similarly.
EXAbåPLE 5
Find log 3 Tii
SOLUTION : By Law 3. we have log3 (27) 1 / 5 1/5) . log 327 .
Since log 3 27 = 3 , we have
1/5 .10g3 27 = 1/5 . 3 = 3/5 .
118
EXAMPLE 6
Assuming x , y and z are positive, express the following in terms of
logarithms of x , y and z .1/7
log b
1/7
SOLUTION : we have log b - (1/7) . log
— (1/7)[ log b y— ( 6.10g b x +4.10g b z)] = (1/7)log b y— (6/7)log b x— (4/7)log b z .
EXAMPLE 7
Solve the following equations :
a.) log 5 3.10g 5 15 —2.10g5 6+ log 5 4— log 5 25
b.) log 6 (3x— 1) — log 6 x = log 64
SOLUTION :
a.) The equation is equivalent to
log 5 (4x + 7) = log 5 15 3 —log 5 6 2 + log 5 4— log 5 25 . (see Law 3.)
Rewriting the equation (using Law 1. and 2.), we obtain
153.4log 5 (4x + 7) = log 5
36.25
hence 4x 15
Checking the solution in the original equation shows that x = 2 is a solution.
b.) log 6 (3x — 1) — log 6 x = log 64
Rewriting the equation using Law 1. and 2. ,we obtain
3x-1log 6 = log 64,
x
3x-1hence
x
We obtain 1
Hence,
This value does not satisfy the original equation since log ( -1) does not exist. So the
equation has no solution.
119
EXERCISES
Solve the following equations :
1. log 7 (3x —5) = 2.10g7 12 -- log 79
2. log 2 (x + 8) — log 2 (2x + 1) = log 8 2 + log 8 4
3. log 13 (x —3) + log 13 (x + 1) = log 13 5
4. 2.10g 2 5 + log 2 4= 2.10g 2 2x
5. log 11 (x + 1) = log 11 4— log Il (x— 2)
6. log 9 (x + 3) + log 9 (x —2) = 1/3.10g 9 8
7. log 10 x + log 10 (4x + 3) = 2.10g 10 5.Jä - log 10 5
8. log (x + I) + log 3 (x— l) = 1
9. h . log b X— log b 4 = log b 3
10. 2 -10.2 12
11. log 8 (x + 2) = log 8 (log 2 8) + log 8 (x—4)
12. 1/3 . log 20 (2x — 1) = log 20 11—2. log 20
13. log 6 (log 6 x) = 0
14. log 4 [ log 4 (log 4 x)] = 0
15. log 3 [ log 4 (log 5 x) ] = 0
16. Solve each system of equations :
a.) h . log 2 y— log 24
log 3 x + 102 3 2y=3
c.) log 2 [log 3 (x + y)]
log x + log y = 3.10g 2
e.) log (x +5)
3x +2y=30
120
+ log 5 ( log 7 7)
b.)
d.)
f.)
log 2 x + log 2 y = 5
log 2 x— log 2 y = 3
log x— h. log y = 0
x 2 —5 y 2 +4=0
log 7 (5x — 3y) = 0
9 x .3 X -
Y =27
Common and natural logarithms
Two bases are commonly employed for logarithms, 10 and e .
a.) Logarithms to the base 10 are called COMMON LOGARITHMS.
Common logarithms are denoted simply as log x or lg x.
That is
log x = lgx = log 10 x.
We can find the common logarithm of a number in a table of logarithm or on a
calculator.
For example,
log 87,6 = 1,9425...
The common logarithm of a number has two parts:
The integer part, called the CHARACTERISTIC of the logarithm. In our examplethe characteristic is 1 .
The other part is called the MANTISSA of the number. The mantissa is always lessthan 1 . In our example the mantissa is 0,9425...
If we are given log x , we can determine the ANTILOGARITI-n.'1 : x .
For example, Find x if log x = 0,444 ,
antilog = 2,78
b.) Logarithms to the base e are called NATURAL LOGARITffv1S .
They are given the special notation In x . That is
In x = log ex.
The base e is irrational , which correct to six places, is
e = 2,71828
Natural logarithms play an important role in both pure and applied mathematics.
Changing the base of a logarithm
Any positive number other than I may be used as a base of a system of logarithms.
A relationship between the logarithms of a number in two different bases will be
developed.
121
By the definition of logarithms
b log x
Taking the logarithm of both sides to base c , we have
log b x. log b = log c x
Thus, we have
log b x
For example,
log x =
logc x
logc b
orIn 10
log xInx =log e
These are the formulas to express natural logarithms in terms of common logarithms or toexpress common logarithms in terms of natural logarithms, respectively.
E)GRCISES
1. Solve the following equations :
a.) 3.10g 5 x + log 25 x = 7
b.) (log 3 x x ).( log 27 x) =4/3
c.) log 5 x 2 -2.10g 25 x = 3
d.) log 25 [ 1/5.10g3 (2— log 0.5 x =- 1/2
e.) log 3 log 2 x)
f.) log 4 x— log 0.25 x = 4
2. Find the domain of the functions :
log (x— 11) y = log (4 —x 2)
log (x — 2).(x — 3) y = log (x 4+16)
e.) Y = log1
y = logx +16
x 4g.) y=log = log (x2 —x 6)y x
log (x + 8) — log (x + Il) + log x
122
3. Solve the following equations :
a.) logs [4—2.10g 6 (5 —x) ] —1/3
b.) log 2
c.) log (4.5 125. (25 x I 1) -2
log(35 — x3 )d.)
log(5 — x)
e.) log (x— 1) + 1/3 = log 8 (x + 7)
f.) log 2+ log (4 +9) = I + log (2 x 2 +1)
g.) log (x + 3) (3X 2 + 2X + 9) = 2
h.) log 3 x— log x 1
4. Write in terms of simpler logarithmic form
a.c b a.blog + log — -2.10gc + log a— log
b
5. log (x) — log (y) = log (x + y) is ft-ue only if :
b.)
c.) x = 0 andy=l
2
6. Solve the equation :
2 6-3
2 —log3 x
123