11 ALGEBRA

2. Types of Logarithm

1. Common Logarithms

Our counting system is based on the number 10. For this reason, a lot of logarithmic work uses the base 10. Logarithms to the base 10 are called common logarithms. We often write log x or lg x to mean log10 x. In this module, we will use log x to mean log10 x.

Common logarithms are widely used in computation. Mathematicians have compiled extensive and highly accurate tables of common logarithms for use in these calculations. These tables and their use will be discussed later in this module.

	2. Natural Logarithms
	Logarithms to the base e are called natural logarithms or Euler logarithms. We often write
e = 2.71828...	ln x to mean the natural logarithm logex.

Natural logarithms are widely used in mathematical analysis in the study of limits, derivatives and integrals.

3. Properties of Logarithms

Property 1 If the argument and the base of a logarithm are equal, the logarithm is equal to 1. Conversely, if the logarithm is 1 then the argument and the base are equal:

		a = b loga b = 1			(a 0,				a 1).
Proof
	By the fundamental identity of logarithms we have aloga N = N. Setting N = a gives us
	aloga a = a = a1, which gives us loga a = 1.							1
	For example, log 3 = 1, log 10 = 1, ln e = 1 and				log				=1.
	3						1	2
							2
	The logarithm of 1 to any base is zero:
Property 2
						.
			loga 1 = 0
Proof	aloga 1 = 1 = a0. So aloga 1 = a0, which gives us loga 1 = 0.
	For example, log3 1 = 0, log 1 1= 0 and log 1 = 0.
	2

Exponential and Logarithmic Functions

149

(x, y 0).

Property 3

Proof

Be careful!

loga(x + y) logax + logay

The logarithm of the product of two or more positive numbers to a given base is equal to the sum of the logarithms of the numbers to that base:

loga(x y) = loga x + loga y

aloga(x y) = x y. Substituting x = aloga x and y = aloga y gives us

aloga(x y) = aloga x aloga y = aloga x + loga y.

Comparing the exponents of the expressions on both sides gives us the required equation: loga(x y) = loga x + loga y.

For example,

log2 6 = log2(2 3) = log2 2 + log2 3 = 1 + log2 3

log3 30 = log3(3 10) = log3 3 + log3 10 = 1 + log3 10 log3 30 = log3(6 5) = log3 6 + log3 5

log2 5 + log2 3 = log2(5 3) = log2 15.

Notice that we can generalize this property as follows:

loga(x1 x2 x3 ... xk) = loga x1 + loga x2 + ... + loga xk (x1, x2, x3, ... , xk > 0).

For example, we can write

log2 30 = log2(2 3 5) = log2 2 + log2 3 + log2 5 = 1 + log2 3 + log2 5.

EXAMPLE

34

Calculate log4

2 + log4 8.

Solution log4 2 + log4 8 = log4(2 8) = log4(4 4) = log4 4 + log4 4 = 1 + 1 = 2

	35									1
EXAMPLE
		Calculate log2 3 + log2 5 + log2										.
											15
Solution		log		3+ log		5+ log			1	= log (3		5		1	)= log		1= 0
								15					15
			2		2		2			2						2

150	Algebra 11

Property 4	The logarithm of the power of a positive number is equal to the product of the power and the
	logarithm of the number.
	loga(xm) = m loga x	(m , x 0).	Be careful!
			(logax)m m loga x

Proof

(am)n = am n

xm = aloga(xm). After substituting x = aloga x on the left side, we get (aloga x)m = aloga(xm), which gives us am loga x = aloga(xm). Since the bases are the same on both sides, we can conclude m loga x = loga(xm).

For example,

log2		8 = log 2(23 )= 3 log 2 2 = 3 1= 3
log			1	= log		1	= log (3 5)= – 5	log		3 = 5 1= 5
		243				35
	3				3		3		3
							3	3 log 2 5.
log2			125 = log 2				53 = log 2 (5 2 )=
								2

Note

n

This property gives us the following special cases:

m xn = xm

1

4b. loga m xn =

n

log a x.

1

= x

– n

4a. log

a

= n log

a

x

x

n

xn

m

EXAMPLE 36

Solution

Write each sum as a single logarithm.						b. ( 1					b) ( 3
a. (2 log	a) + (3 log	3	b) – log	3	c		log	2	a)+(3 log	2		log	2	c)
3						2					2

We apply the property loga(xm) = m loga x.

a. (2 log3 a) + (3 log3 b) – log3 c = log3(a2) + log3(b3) + (–1) log3 c log3(a2) + log3(b3) + log3(c–1) = log3(a2 b3 c–1) = log3( a2 cb3 )

	( 1	log2 a)+(3					log 2 b) (				3	log 2 c)= log 2					1	+ log 2 b3+(			3)	log 2 c
b.																a2
	2										2										2
										3									1
	log			a + log		b3 + log				c( 2 ) = log					a+ log		b3 + log (			)
		2			2				2				2			2
																		2	c3
	log2		(	a b3			1	)= log 2(				a b3		)
							c3					c3

Exponential and Logarithmic Functions

151

EXAMPLE	37	Calculate log2	4 2	8 3 16 .

Solution

Property 5

Proof

am = am – n

an

			4 2			1			4 2				1
log2 4 2	8 3 16 = log 2			8 16		3 = log 2				2 3 (2 4) 3
				4						13				13	1		13
	= log2		4 2	23 3	= log 2		4	2		2 3	= log 2		4	2 (2 3 ) 2 = log 2		4	2 1+ 6
			19			19		19				19
	= log		4 2 6	= log		(2 24 )=				log	2 =
		2			2			24		2		24
										1

The logarithm of the quotient of two positive numbers is equal to the difference between the

logarithms of the dividend and the divisor to the same base:

x

Be careful!

loga( y)= log a x log a y .

loga x

loga x – loga y

loga y

x

x

= a

loga y

log

x

log

y

on the left side, we obtain

y

. If we substitute x = a

a

and y = a

a

x

x

( x)= log

aloga x = aloga ( y)

aloga x loga y = aloga ( y)

log

a

a

x log

a

y.

aloga y

y

For example,

log

5 = log

5 log

3

2 3

2

2

log

(0.12)= log (

12 )= log (

3 )= log

3

log (5

2)= log

3 – 2

5

5

100

5

25

5

5

5

40 = log

log

10 + log

4 – log

5 = log (10 4) – log

5 = log

40 – log 5 = log

2

2

8 = 3.

2

2

2

2

2

2

2

5

Notice that we can combine properties 4 and 5 to write expressions with addition and subtraction of logarithms as the logarithm of a single fraction. The addends form the numerator of the fraction and the subtrahends form the denominator, for example:

log

b + log

c – log

d + log

e – log f

= log

(

b c e

).

a

a

a

a

a

a

d f

As a numerical example, consider

log3 15 – log3 5 + log3 6 – log3 2 = log3(155 26) = log3 9 = log3(32) = 2. Remember that this property only applies to logarithms with a common base.

152	Algebra 11

EXAMPLE 38

Solution

EXAMPLE 39

Solution

log 10 = log1010 = 1

EXAMPLE 40

Solution

Property 6

–

Express log 30 and log 3.3 in terms of p given log 3 = p.

Since 30 = 3 10, we get log 30 = log(3 10) = log 3+ log 10 = p+1.

p 1

Since 3.3 = 103 , we have log 3.3 = log 103 = log 10 log 3 =1 p.

Given log 300 = 2.47712, calculate log(0.0027).

log(0.0027) = log(10274 ) = log 27 – log 104 = log 33 – 4 log 10

= (3 log 3) – 4 (1)

log 300 = log(3 100) = log 3 + log 102 = log 3 + 2 log 10 = 2 + log 3. So log 3 = log 300 – 2. Using log 300 = 2.47712, we get log 3 = 2.47712 – 2 = 0.47712. (2)

Combining (1) and (2) gives us log(0.0027) = (3 0.47712) – 4 = – 2.56864.

Write each logarithm as a sum or difference of logarithms to base a.

a. loga b3c2 d4e5

a.loga( b3 c2 ) d4 e5

b.We have loga

5(b+ c)2 b. loga (d e)3

=loga b3 + loga c2 – loga d4 – loga e5 = 3 loga b + 2 loga c – 4 loga d – 5 loga e

5	(b+ c)	2					2
			= log		5 (b+ c)2 log		(d e)3 = log ( b+ c) 5	3log ( d e)
(d e)3
				a		a	a	a

= 25 loga(b+ c) 3log a(d e).

Notice that logarithms cannot be distributed over addition or subtraction, and also that logarithms enable us to perform simpler operations (addition and subtraction) instead of multiplication and division. This is why logarithms are so useful in computation.

1. Changing the Base of a Logarithm

Most of the properties that we have seen so far only apply to logarithms to the same base. Now we will consider some key properties which allow us to deal with logarithms to different bases.

Raising the base of a logarithm to a non-zero power is the same as dividing the logarithm by that power:

logan x = n1 log a x .

Exponential and Logarithmic Functions

153

Proof

EXAMPLE 41

x = aloga x and x = (an)logan x = anlogan x, so aloga x = anlogan x.

Since the bases are the same, we can equalize the exponents: loga x = n logan x.

Hence log

n x =

1

log

x, as required.

a

a

n

1 log

1

For example, log

3 = log

2 3 =

3 and

log

9 = log

–1 9 =

log

32

= –2.

4

2

1

3

3

2

2

–1

3

Write the following expression as a single logarithm to base 3.

	log 1	7+2 log9 49	log		1
				3	7
	3

Solution Expressing the bases as powers of 3, we get

log 1	7+2 log 2	49 log 1/ 2	1	=	1	log 3 7+2		1	log 3 49		1	log 3	1
			7					2			1		7
3	3	3			1
											2

–log3 7 + log3 72 – 2log3 7–1 = –log3 7 + 2log3 7 + 2log 7 = 3log3 7 = log3 343.

Note

As a result of property 6,

6a.

n xm = m

log

.

6b.

.

log

a

x

log

a

x = log

n xn

a

n

a

Check Yourself 7

Evaluate the expressions.

a.	log36	1	+ log 1	1		+ log 8	128+ log 1	9	b.	8	log 3	1+log81 4
											2	c. 7293
		6	5	125			3
Answers
a.	17		b. 729				c. 72
	6

The following property gives us a another rule for changing the base of a logarithm.

Property 7 Let a, b and x be positive numbers such that a 1 and b 1. Then

loga x = logb x . logb a

This formula is called the Change of Base formula.

154	Algebra 11

Proof

EXAMPLE 42

Solution

Let loga x = y. By definition, ay = x.

So x = blogb x and a = blogb a.

If we substitute these in ay = x, we get

(blogb a)y = blogb x by logb a = blogb x.

Since the bases are the same on both sides, we can equalize the exponents:

y log

a = log

x y =

logb x

.

b

b

logb a

Since log

x = y, we conclude log

a

x =

logb x

, as required.

a

logb a

For example,

log2 7 =

log3 7

= log 7

= ln7

, log 3

5 =

log5 5

=

1

,

log3 2

log 5 3

log 2

ln2

log 5 3

log 5 =

log2 5

and ln x=

log3 x

.

log2 10

log3 e

Scientific calculators do not have a

button but they do have a

button (for base

logx y

log

10 logarithms) and an

button (for natural logarithms). To calculate logarithms to a different

ln

base on a calculator, we therefore use the Change of Base

Example: Find the value of log27 using a scientific calculator

0.84509804

7

log

0.30102999

/

2

log

=2.80735492

Conclusion

We can easily derive the following properties from the examples we have studied:

7a.

loga b

=

1

and

loga b logb a = 1

for a, b > 0 and a, b 1.

logb

a

7b.

loga x

=

logb x

for a, b, x, y > 0 and a, b 1.

loga y

logb y

Evaluate

4

+

2

.

log2 12

log3 12

4

+

2

= 4

1

+2

1

= 4 log12 2+ 2 log12 3

log2 12

log 3 12

log 2

12

log 3 12

= log

12

24

+ log

12

32 = log

(2 4 32)= log

12

12 2

= 2.

12

Exponential and Logarithmic Functions

155

EXAMPLE 43

Solution

EXAMPLE 44

Solution

EXAMPLE 45

Solution

Calculate log20 200 in terms of p if log5 2 = p.

We begin by changing log20200 to base 5:

log20 200 =

log 200

=

log (8 25)

=

log

8+ log

25

=

log 2 3+ log

5 2

=

3 log 2+2

log5 20

log5(4 5)

log 5 4+ log 5 5

log5 22

+1

2 log5 2+1 .

5

5

5

5

5

5

5

Using the substitution log52 = p, we obtain log20

200 =

3p+2.

2p+1

loga b = c is given. Express each logarithm in terms of c.

a. log 2	b	ab2	b.	log	a	3	ab4
a
					b2

In each case we use the Change of Base formula.

a.

log 2

ab2

=

log

a

ab2

=

log

a

a+ log

a

b2

=

1+2 log

a

b

=

1+2c

loga a2b

loga a2 + log a b

2 loga a+ c

2+ c

a

b

b.

log

ab4 =

log

a

ab4

=

log

a

a+ log

a

b4

=

1+4 log

a

b

1+4c

3

a

3

loga a

3

loga b

2

3loga a 2 log a b

3 2c

a

2

b

loga b2

Calculate each logarithm in terms of a, using the relation given.

a. log 25; a = log 2		b. log3 18; a = log3 12
c. log12 27; a = log6 16
a. log 25 = log 52 = 2log 5 = 2 log	10	= 2(log 10 – log 2) = 2(1 – a) = 2 – 2a
	2

b.log3 18 = log3(32 2) = log3 32 + log3 2 = 2 + log3 2. So we need to find log3 2 by using log3 12 = a:

log3 12 = log3(22 3) = log3 22 + log3 3 = 2log3 2 + 1. So 2log3 2 + 1 = a, i.e. log3 2 = a2– 1.

In conclusion we can write log3 18 = 2+ log 3	2 = 2+ a – 1= a+3.
	2	2

156	Algebra 11

c.

log

12

27 =

log2 27

=

log 2 33

=

3log 2 3

=

3log 2 3

(1)

log

2 12

log2(22

3)

log 2 22 + log

2 3

2+ log 2

3

log6 16 =

log2

16

=

log2 24

=

4

=

4

log2 6

log2(3 2)

log 2 3+ log 2 2

log 2 3+1

Using the equality log6 16 =

4

= a, we get log2 3 =

4 a .

(2)

log2 3+1

a

3(

4 a

)

12 3a

Substituting (2) into (1) gives us log12 27 =

a

=

.

4 a

a+4

2+(

)

a

46 Simplify

1

1

EXAMPLE

25log6 5 +49log8 7 .

Solution

1

1

25log6 5 +49log 8 7

=

25log 5 6 +49log7 8

=

52 log 5 6 +72 log7 8 =

5log 5 6 2 +7log7 82

= 6 2 +8 2 =10

aloga x = x

47

Show that log2 3 log5 7

log11 13

= log2 13 log5 3

log11 7.

EXAMPLE

Solution If we use the Change of Base formula on the left-hand side with any base a 1, we get

log2 3 log5 7

log1113 =

loga 3

log a 7

log a 13

=

log a13

log a 3

log a7

loga 2

loga 5

log a 11

log a 2

log a 5

log a11

= log2 13

log5 3 log11 7.

Check Yourself 8

1.

Evaluate the expressions.

log3 135

log 3 45

a. log3 2

log4 3 log5 4 log6 5

log7

6 log8 7

b.

log75 3

log675

3

2. Calculate each logarithm in terms of the variable(s) provided, using the given relation(s).

a. log 40;

log 2 = a

b. log 45; log 2 = a and log 3 = b

c. log3 5;

log6 2 = a and log6 5 = b

d. log100 40;

log2 5 = a

e. log275 60; log12 5 = a and log12 11 = b

f.

log6 16;

log12 27 = a

Exponential and Logarithmic Functions

157

3. Prove each statement by using the properties of logarithms.

	a.	logbn(an)=		logb a+ logb n		b.	loga n		=1+ log a b
				1+ logb n			logab n
2006! =	d.	1		1	1
2006 2005 2004 ... 2 1		log2 N	+ log3 N +...+ log 2006 N =1 if N = 2006!
Answers
1.	a.	1	b.	–3
		3
2.	a.	2a + 1	b.	1 – a + 2b	c.	b		d.		3+ a
						1– a				2(a+1)

c. logab n =

e. 1+ a 2a+ b

loga n logb n

loga n+ log b n

f. 4(3 – a)

3+ a

Property 8		bloga c = cloga b	for a, b, c 0 and a 1.

Proof bloga c = (aloga b)loga c = (aloga c)loga b

= cloga b because aloga c = c.

So we obtain bloga c = cloga b.

(am )n

=(an )m

For example, 7log2 5 = 5log 2 7 and 3log 2 = 2log 3.

EXAMPLE

48 Calculate 2log3 5 – 5log3 2.

Solution By property 8, since 2log3 5 = 5log3 2, we have 2log3 5 – 5log3 2 = 5log3 2 – 5log3 2 = 0.

EXAMPLE

49 Calculate (2log 27)log3 5 8log 2.

Solution

Using 2log 27 = 2log 33 = (23)log 3

= 8log 3 = 3log 8 and log 5 = log

3

10

= log

3

10 – log

3

2,

3

2

we can write (2log 27)log3 5 8log

2

= (3log 8)log3 10 – log3 2 8log 2

log 8

log3 10

(a

m

n

n

)

m

=

(3 log 8) log

2

8log 2

)

=(a

(3

)

3

3log3 10 =10

(3log3 10 )log 8

10log 8

3log3 2 = 2

=

8

log 2

=

log 2

= 8.

10

log 8

= 8

(3

log

2

)

log 8

8

3

2log 8

8log 2

= 2log 8

158	Algebra 11