skill

S k i l l
i n
A L G E B R A

LOGARITHMS

Definition

Common logarithms

The three laws of logarithms

WHEN WE ARE GIVEN the base 2, for example, and exponent 3, then we can evaluate 2³.

2³ = 8.

Inversely, if we are given the base 2 and its power 8 --

2^? = 8

-- then what is the exponent that will produce 8?

That exponent is called a logarithm. We call the exponent 3 the logarithm of 8 with base 2. We write

3 = log₂8.

The base 2 is written as a subscript.

3 is the exponent to which 2 must be raised to produce 8.

A logarithm is an exponent.

Since

10⁴ = 10,000

then

log₁₀10,000 = 4.

"The logarithm of 10,000 with base 10 is 4."

4 is the exponent to which 10 must be raised to produce 10,000.

"10⁴ = 10,000" is called the exponential form.

"log₁₀10,000 = 4" is called the logarithmic form.

Here is the definition:

log_bx = n means bⁿ = x.

That base with that exponent produces x.

Example 1. Write in exponential form: log₂32 = 5

Answer. 2⁵ = 32

Example 2. Write in logarithmic form: 4⁻² =

1
16

Answer. log₄

1
16

= −2.

Problem 1. Which numbers have negative logarithms?

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Do the problem yourself first!

Proper fractions.

Lesson 20 of Arithmetic

Example 3. Evaluate log₈1.

Answer. 8 to what exponent produces 1?

8⁰ = 1.

log₈1 = 0.

We can observe that, in any base, the logarithm of 1 is 0.

log_b1 = 0

Example 4. Evaluate log₅5.

Answer. 5 with what exponent will produce 5? 5¹ = 5. Therefore,

log₅5 = 1.

In any base, the logarithm of the base itself is 1.

log_bb = 1

Example 5. log₂2^m = ?

Answer. 2 with what exponent will produce 2^m ? m, obviously.

log₂2^m = m.

The following is an important formal rule, valid for any base b:

log_bb^x = x

This rule embodies the very meaning of a logarithm. x -- on the right -- is the exponent to which the base b must be raised to produce b^x.

Example 6 . Evaluate log₃

1
9

Answer.

1
9

is equal to 3 with what exponent?

1
9

= 3⁻².

log₃

1
9

log₃3⁻² = −2.

Compare the previous rule.

Example 7. log₂ .25 = ?

Answer. .25 = ¼ = 2⁻². Therefore,

log₂ .25 = log₂2⁻² = −2.

Example 8. log₃ = ?

Answer. = 3^1/5. (Definition of a rational exponent.) Therefore,

log₃ = log₃3^1/5 = 1/5.

Problem 2. Write each of the following in logarithmic form.

a) bⁿ = x.	log_bx = n.	b) 2³ = 8.	log₂8 = 3.

c) 10² = 100.	log₁₀100 = 2.	d) 5⁻² = 1/25.	log₅1/25 = −2.

Problem 3. Write each of the following in exponential form.

a) log_bx = n.	bⁿ = x.	b) log₂32 = 5.	2⁵ = 32.

c) 2 = log₈64.	8² = 64.	d) log₆1/36 = −2.	6⁻² = 1/36.

Problem 4. Evaluate the following.

a) log₂16	= 4	b) log₄16	= 2

c) log₅125	= 3	d) log₈1	= 0

e) log₈8	= 1	f) log₁₀1	= 0

Problem 5. What number is n?

a) log₁₀n = 3.	1000	b) 5 = log₂n.	32

c) log₂n = 0.	1	d) 1 = log₁₀n.	10

e) log_n	1 16	= −2.	4	f) log_n	1 5	= −1.	5

g) log₂	1 32	= n.	−5	h) log₂	1 2	= n.	−1

Problem 6. log_bb^x = x

Problem 7. Evaluate the following.

a) log₉

1
9

b) log₉	1 81	= −2	c) log₂	1 4	= −2

d) log₂	1 8	= −3	e) log₂	1 16	= −4

f) log₁₀.01	= −2	g) log₁₀.001	= −3

h) log₆	= 1/3

Write in exponential form, and apply Problem 6.

i) log_b

Common logarithms

The system of common logarithms has 10 as its base. When the base is not indicated:

log 100 = 2

then the system of common logarithms -- base 10 -- is implied.

Here are the powers of 10 and their logarithms:

Powers of 10:	1 1000	1 100	1 10	1	10	100	1000	10,000

Logarithms:	−3	−2	−1	0	1	2	3	4

Logarithms replace a geometric series with an arithmetic series.

Problem 8.

a) log 10⁵ = 5. 10 is the base.

b) log 10ⁿ = n

c) log 58 = 1.7634. Therefore, 10^1.7634 = 58

Problem 9. log (log x) = 1. What number is x?

(See above.)

x = 10¹⁰ = 10,000,000,000

The three laws of logarithms

1. log_bxy = log_bx + log_by

"The logarithm of a product is equal to the sum
of the logarithms of each factor."

2. log_b

= log_bx − log_by

"The logarithm of a quotient is equal to the logarithm of the numerator
minus the logarithm of the denominator."

3. log_b x ⁿ = n log_bx

"The logarithm of a power of x is equal to the exponent of that power
times the logarithm of x."

For a proof of these laws, see Topic 20 of Precalculus.

Example 9. Apply the laws of logarithms to log

abc²
d ³

Answer. According to the first two laws,

log	abc² d ³	=	log (abc²) − log d ³

		=	log a + log b + log c² − log d ³

		=	log a + log b + 2 log c − 3 log d,

according to the third law.

The Answer above shows the complete theoretical steps. In practice, however, it is not necessary to write the line

log

abc²
d ³

log (abc²) − log d ³

The student should be able to go immediately to the next line --

log

abc²
d ³

log a + log b + log c² − log d ³

-- if not to the very last line

log abc²
d ³ = log a + log b + 2 log c − 3 log d.

Example 10. Apply the laws of logarithms to log
z⁵ .

Answer. log
z⁵ = log x + log − log z⁵

Now, = y^½. (Lesson 29.) Therefore, according to the third law,

log

z⁵

= log x + ½ log y − 5 log z.

Example 11. Use the laws of logarithms to rewrite log (sin x log x)

Solution. This has the form log ab. a = sin x, b = log x. Therefore,

log (sin x log x) = log sin x + log log x.

Example 12. Use the laws of logarithms to rewrite log .

Solution.

log	=	log (x cos x)

	=	½ log (x cos x), 3rd Law

	=	½ (log x + log cos x), 1st Law.

Problem 10. Use the laws of logarithms to rewrite the following.

a) log	ab c	= log a + log b − log c

b) log	ab² c⁴	= log a + 2 log b − 4 log c

c) log	z	= 1/3 log x + 1/2 log y − log z

d) log (sin²x log x)	=	log sin²x + log log x

	=	2 log sin x + log log x

e) log	=	log (sin x cos x)^1/2

	=	½ log (sin x cos x)

	=	½ (log sin x + log cos x).

Example 13. Given: log 3 = .4771 Evaluate

a) log 3000

Solution. Write 3000 in scientific notation:

log 3000	=	log (3 × 10³)

	=	log 3 + log 10³

	=	.4771 + 3

	=	3.4771

b) log .003

Solution.	log .003	=	log (3 × 10⁻³)

		=	log 3 + log 10⁻³

		=	.4771 − 3

		=	−2.5229

Problem 11. Given: log 6 = .7781 Use the laws of logarithms to evaluate the following.

a) log 600	=	log (6 × 10²)

	=	log 6 + log 10²

	=	.7781 + 2

	=	2.7781

b) log 60	=	log (6 × 10)

	=	log 6 + log 10

	=	.7781 + 1

	=	1.7781

c) log .06	=	log (6 × 10⁻²)

	=	log 6 + log 10⁻²

	=	.7781 − 2

	=	−1.2219

Example 14. Given: log 2 = .3010, log 3 = .4771 Evaluate log 18.

Solution. 18 = 2· 3². Therefore,

log 18	=	log (2· 3²)

	=	log 2 + log 3²

	=	log 2 + 2 log 3

	=	.3010 + 2(.4771)

	=	.3010 + .9542

	=	1.2552

Problem 12. Given: log 2 = .3010 log 3 = .4771 log 5 = .6990

Use the laws of logarithms to find the following.

a) log 6 = log 2 + log 3 = .7781

b) log 15 = log 3 + log 5 = 1.1761

c) log 4 = log 2² = 2 log 2 = .6020

d) log 8 = log 2³ = 3 log 2 = .9030

e) log 30 = log 3 + log 10 = 1.4771

f) log 300 = log 3 + log 100 = 2.4771

g) log 3000 = log 3 + log 1000 = 3.4771

h) log 12 = log 3 + log 4 = 1.0791

i) log

3
5

j) log = ½ log 3 = .2386

k) log = ½ log 5 = .3495

l) log

m) log

n) log = ½(log 2 − log 3) = −.0881

o) log 1500 = log 3 + log 5 + log 100 = 3.1761

For the system of natural logarithms, see Topic 20 of Precalculus.

For logarithmic and exponential "functions," see Topic 21.

Next Lesson: Variation

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