bA = log Ax =

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Instruction: Logarithms

In the equation b^x =A, we call x an exponent, b a base, and A the x-th power of base b.

We also call x the logarithm of A to the base b, which we write as log_b A=x. (We understand the base to be a positive number not equal to one.)

In words, the definition of a logarithm states, “The logarithm of a number to any base is the exponent of that power of the base.” The definition gains clarity with symbols.

A logarithm then is the exponent applied to some fixed number called a base in order to obtain a specified power of the base called the argument of the logarithm. In the box above, A is both the power of base b and the argument of the logarithm. Likewise, x is both the exponent applied to b that obtains A and the logarithm to base b.

Mathematicians use special notation for common logarithms—logarithms (base 10) — and for natural logarithms—logarithms (base e). Usually, subscript to the word “log” denotes the logarithm's base. With common logarithms, the subscript is missing and the base is simply understood to be ten. Thus, log100 reads, "logarithm (base ten) of one hundred." With natural logarithms, the abbreviation “LN” or “ln” replaces the word “log,” and the base is understood to be e. Thus, ln reads, "logarithm (base e) of two" or "natural log of two." 2

Since logarithms are exponents, evaluating logarithms requires students to be familiar with the exponent properties in the box below.

Evaluating logarithms is analogous to evaluating exponential expressions. It is easy to mentally evaluate 2 as 32 or even ⁵ 8^{2 3} as 4, but we use calculators to evaluate an expression like 2 or ^{1 2} e³. Similarly, evaluating logarithmic expressions sometimes requires a calculator.

For introductory purposes, this lecture will evaluate logarithms that do not require a calculator.

If ,a b≠0, then

( )

^p

p r r p r

a = a = a ,

0 1

a = ,

1 ^r 1

r

a r

a a

− =⎛ ⎞⎜ ⎟⎝ ⎠ = ,

r r r

r

a b b

b a a

⎛ ⎞− =⎛ ⎞ =

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ . Definition of Logarithm:

If b^x = A then

log

_b

A = x

where b is a base, that is, b>0, 1b≠ .

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For example, consider log 9 . This expression reads, "logarithm (base three) of nine." The ₃ logarithm equals two because nine is the second power of base three. More examples appear below.

Example 1

The argument of the log is forty-nine, and the base is seven. The reader should recognize forty- nine as the second power of seven. Hence, log7

( )

49 =2, i.e., since 7² =49, log7

( )

49 =2.

Example 2

The argument of the log is eighty-one, and the base is three. The reader should recognize eighty- one as the fourth power of three. Hence, log 813

( )

=4, i.e., since 3⁴ =81, log 813

( )

=4.

Example 3

The argument of the log is one, and the base is pi. The reader should recognize one as the zero power of any non-zero base. Hence, ^logπ

( )

¹ =⁰, i.e., since π⁰ =1, ^logπ

( )

¹ =⁰.

Example 4

The argument of the log is eight, and the base is two. The reader should recognize eight as the third power of two. Hence, log2

( )

8 =3, i.e., since 2³ =8, log2

( )

8 =3.

Example 5

The argument of the log is sixty-four, and the base is two. The reader should recognize sixty- four as the sixth power of two. Hence, log2

( )

64 =6, i.e., since 2⁶ =64, log2

( )

64 =6.

Evaluate log7

( )

49 without using a calculator.

Evaluate log 813

( )

without using a calculator.

Evaluate ^logπ

( )

¹ without using a calculator.

Evaluate log2

( )

Evaluate log2

( )

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Example 6

The argument of the log is two, and the base is four. The reader should recognize two as the square-root of four. The square-root of four is also the one-half power of four. Hence,

( )

log4 2 =1 2. In other words, since 4=4^{1 2} =2,log4

( )

2 =1 2. Example 7

The argument of the log is 0.01, and the base is ten. The reader should recognize 0.01 as 1 100, which is the negative second power of ten. Hence, ^{log 0.01}

( )

^{= −}². In other words, since

2 2

1 1

10 0.01

10 100

− = = = , ^{log 0.01}

( )

^{= −}²^.

Example 8

The argument of the log is the irrational number e⁶. The notation “ln” refers to the natural log, which is the logarithm to base e, so the base of the logarithm is e. The reader should recognize

e6 as the sixth power of e. Hence, ^ln

( )

^e⁶ ⁼⁶^.

Example 9

The argument of the log is the fraction four-ninths, and the base is two-thirds. The reader should recognize four-ninths as the second power of two-thirds. Hence, log2 3

( )

4 9 =2, i.e., since

2 2 4

3 9

⎛ ⎞ =

⎜ ⎟⎝ ⎠ , then ₂

3

log 4 2

⎛ ⎞ =9

⎜ ⎟⎝ ⎠ .

Evaluate log4

( )

Evaluate ^{log 0.01}

( )

without using a calculator.

Evaluate ^ln

( )

^e⁶ without using a calculator.

Evaluate log2 3

( )

4 9 without using a calculator.

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Instruction: Logarithm Properties The previous lecture defined a logarithm as below:

From this definition, the following two properties of logarithms are apparent.

The remainder of this lecture states and proves four other logarithm properties. We refer to the first of these four properties of logarithms as the Sum of Logs Theorem.

Proof: Let log_bM =x and log_b N = y. Then, by definition, we have b^x =M and b^y =N. Consequently, M N⋅ =b b^x⋅ ^y =b^{x y}⁺ . Then, by definition, we obtain ^logb

(

M N⋅

)

= +x y; whence, by substitution, we acquire, ^logb

(

M N⋅

)

=^logb

( )

M +^logb

( )

N

We refer to the next property of logarithms as the Difference of Logs Theorem.

Proof: Let log_bM =x and log_b N = y. Then, by definition, we have b^x =M and b^y =N. Consequently, M N =b b^x ^y =b^{x y}⁻ . Then, by definition, we obtain ^logb

(

M N

)

= −x y; whence, by substitution, we acquire, ^logb

(

M N

)

=^logb

( )

M −^logb

( )

N .

We refer to the next property of logarithms as the Log Multiple Theorem.

Sum of Logs Theorem: The logarithm of a product, to any given base, equals the sum of the logarithms, i.e.,

(

^⋅

)

⁼

( )

⁺

( )

log_b M N log_b M log_b N

where M and N are positive real numbers, and b is a positive number not equal to one.

Difference of Logs Theorem: The logarithm of a ratio, to any given base, equals the logarithm of the numerator minus the logarithm of the denominator, i.e.,

( ) ( )

⎛ ⎞ = −

⎜ ⎟

⎝ ⎠

log_b M log_b log_b

M N

N

where M and N are positive real numbers, and b is a positive real number ∋ ≠b 1.

Log Multiple Theorem: The logarithm of a power of a number, to any given base, equals the logarithm of a number multiplied by the exponent of the power, i.e.,

( )

^{= ⋅}

^{( )}

log_b M^P P log_b M

where M is a positive real number, b is a positive real number not equal to one, and P is any real number.

If b^x =A where b>0 and b≠1, then log_b A=x.

( )

log_b b^P =P and b^log^b^P =P

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Proof: Let log_bM =x. By definition, b^x =M . Raising M to the P, we have ^M^P ⁼

( )

^b^x ^P ⁼^b^{P x}^⋅ ^.

Since b^{P x}^⋅ =M^P, then, by definition, we obtain ^log^b

( )

^M^P ^{= ⋅}^{P x}; whence, by substitution, we acquire, ^log^b

( )

^M^P ^{= ⋅}^P ^log^b^M^.

We refer to our final property of logarithms as the Change of Base Theorem.

Proof: Let ^logβ

( )

^A =^x, ^logβ

( )

^b = ^y, and ^logb

( )

A =z. Then, by definition, we obtain the following three equations.

) )

)

x y z

i A

ii b

iii b A β β

=

Using equations i and iii, we have β^x =b^z by the transitive property. Using equation ii, we have

( )

^z

x y y z

β = β =β ^⋅ by substitution. Hence, x=zy becauseβ^x=β^{y z}^⋅ . Substituting for x, y, and z, we acquire: ^log_β

( )

A =^logb

( )

A ⋅^log_β

( )

b . Division completes the proof as shown below.

( ) ( )

log log

b A b

A b

β β β

= ⋅

( )

log_β b

( ) ( ) ( )

log log

log ^b

A A

b

β β

=

The table on the following page demonstrates the application of the preceding theorems and properties.

Change of Base Theorem: A defined ratio of logarithms in any given base can be written as a single logarithm of the ratio’s numerator to any other given base, i.e.,

( ) ( ) ( )

β β

log =

log log^b

A A

b

where A, b, and β are positive real numbers ∋ ≠b 1,β ≠1.

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Logarithm Properties Example 1 Example 2

( ) ( ) ( )

log_b MN =log_b M +log_b N ^log^x⁺^log

(

^x^{+ =}¹

)

^{log (}

[

^{x x}⁺¹⁾

]

^log2¹⁶=^log2⁴+^log2⁴

( ) ( ) ( )

log_b M N =log_b M −log_b N

(

x

)

x

x 1 ln 1 ln

ln ⎟= + −

⎠

⎜ ⎞

⎝

⎛ +

2 2 2

log log log x

x y

y

− = ⎛ ⎞⎜ ⎟

⎝ ⎠

( ) ^{( )}

log_b M^p = ⋅p log_b M _b x log_b x

2

log = 1⋅ log₂5³ =3⋅log₂5

( ) ( ) ( )

log log

log ^b

A A

b

β β

= b

a a

b ln

log =ln ln 8 ₂

log 8 3

ln 2 = =

( )

log_b b^p = p log₂2³ =3 lne⁹ =9

log_bA

b = A 2^log²⁶ =6 e^ln⁷ =7

Instruction: Approximating the Natural Logarithm Function

Consider the natural logarithm function f(x)=ln

( )

x . Like all logarithmic functions, f(x) takes rational values only when its argument equals a rational power of the base. The base e, however, is, itself, an irrational number, so one (which is equal to e⁰) is the only integer

argument for which f(x) will take a rational number value. Since ln(x) takes irrational values for integer inputs, a quick method for approximating ln(x) would be useful.

The Log Multiple Theorem states that the log of the p-th power of a number is p times the log of that number, in symbols, log_b A^p = ⋅p log_b A. Consider that ln(3) ≈ 1.1. Note that the natural log of nine equals ln3², which, in turn, equals 2ln3, which is approximately 2(1.1) or 2.2.

Accordingly, various values of ln(x) can approximated as multiples of 1.1. For example, ln(27)

= ln3³ = 3ln3 ≈ 3(1.1) ≈ 3.3. Note further that since ln(9) ≈ 2.2 and ln(27) ≈ 3.3 then ln(10), ln(15), and ln(25) all take values between 2.2 and 3.3. Hence, we have a method for

approximating values of the natural log without a calculator. This method takes advantage of the fact that the natural logarithm function is an increasing function that increases very slowly.

Example

Note that eighty-five is close to eighty-one, the fourth power of three. Accordingly, we see

( ) ( ) ( )

⁴

^{( )}

ln 85 ≈ln 81 ≈ln 3 ≈ ⋅4 ln 3 ≈ ⋅4 1.1. Hence, ^{ln 85}

( )

⁼^4.4^.

Mentally approximate ^{ln 85}

( )

^.

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Instruction: Introduction to Logarithms

Example 1

Evaluating Logarithms without a Calculator

Recognize the argument 1 16 as a power of the base 4.

2 2

1 1

16 4 4

= = −

Since 1 ² 16 4

= − , ₄ 1

log 2

16

⎛ ⎞ = −

⎜ ⎟

⎝ ⎠ .

Example 2

Convert the argument 0.0016 to a fraction and reduce.

16 24

0.0016

10, 000

= =

24 ⁴ ⁴

1 5 5

⋅ =

Recognize the argument 1 5 as a power of the base ⁴ 1 5.

4 4

1 1

0.0016

5 5

= = ⎜ ⎟⎛ ⎞⎝ ⎠

Since

1 4

0.0016 5

= ⎜ ⎟⎛ ⎞⎝ ⎠ , 1

( )

5

log 0.0016 =4. Evaluate ₄ 1

log 16

⎛ ⎞

⎜ ⎟

⎝ ⎠ without using a calculator.

Evaluate 1

( )

5

log 0.0016 without using a calculator.

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Example 3

Recognize the argument 4 as a root (fractional power) of the base 64.

1

3 3

4= 64 =64

Since

1

4=643, 64

( )

log 4 1

=3.

Example 4

Recognize the argument 64 and the base 16 as powers of 4.

2 3

4 16

4 64

=

Rewrite the argument as a power of 16.

( )

³ ¹ ³ ³

3 2 2

64 4 16 ⎛16 ⎞ 16

= = =⎜ ⎟ =

⎝ ⎠

Since

3

64=162, then 16

( )

16 ³²

log 64 log 16 3 2

⎛ ⎞

= ⎜ ⎟=

⎝ ⎠ .

Evaluate log64

( )

Evaluate log16

( )

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Instruction: Properties of Logarithms

Example 1

Expanding a Logarithm into the Product of a Number and a Logarithm

Apply the property of logs that states ^log^b

( )

^M^p ^{= ⋅}^p ^log^b

^{( )}

^M ^.

( )

³

^{( )}

log_b a =3log_b a Example 2

Expanding a Logarithm into a Sum or Difference of Logarithms

Apply the property of logs that states ^logb

(

M N

)

=^logb M −^logb N.

( ) ( )

2

2 3

log_b x3 log_b log_b

x y

y

⎛ ⎞

= −

⎜ ⎟

⎝ ⎠

( )

^M^p ^{= ⋅}^p ^log^b

^{( )}

^M ^.

( )

²

( )

³

^{( )} ^{( )}

log_b x −log_b y =2 log_b x −3log_b y Rewrite ^log^b

( )

^a³ as a product of a number and a logarithm.

Expand

2

log_b x3

y

⎛ ⎞

⎜ ⎟

⎝ ⎠ as a sum or difference of logarithms. Remove all exponents from the argument.

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Example 3

( )

^M^p ^{= ⋅}^p ^log^b

^{( )}

^M ^.

2 3 2

log_b a b 3 log_b a b

cd cd

⎡⎛ ⎞ ⎤ ⎛ ⎞

⎢⎜ ⎟ ⎥ = ⋅ ⎜ ⎟

⎢⎝ ⎠ ⎥ ⎝ ⎠

⎣ ⎦

(

M N

)

=^logb

( )

M −^logb

( )

N .

( ) ^{( )}

2

3 log_b a b 3 log_b 2 log_b

a b cd

cd

⎛ ⎞ ⎡ ⎤

⋅ ⎜⎝ ⎟⎠= ⎣ − ⎦

(

MN

)

=^logb

( )

M +^logb

( )

N .

( ) ^{( )} ( ) ^{( )} ( ^{( )} ^{( )} ) ( ) ^{( )} ^{( )} ^{( )}

2 2

2

3 log log 3 log log log log

3 log log log log

b b b b b b

b b b b

a b cd a b c d

a b c d

⎡ − ⎤= ⎡ + − + ⎤

⎣ ⎦ ⎣ ⎦

⎡ ⎤

= ⎣ + − − ⎦

( )

^M^p ^{= ⋅}^p ^log^b

^{( )}

^M ^.

( )

²

^{( )} ^{( )} ^{( )} ^{( )} ^{( )} ^{( )} ^{( )}

3 log⎡⎣ _b a +log_b b −log_b c −log_b d ⎤⎦=3 2 log⎡⎣ _b a +log_b b −log_b c −log_b d ⎤⎦ Apply the property of logs that states log_bb^p = p.

( ) ( ) ( ) ( ) ( ) ( ) ( )

3 2 log⎡⎣ _b a +log_b b −log_b c −log_b d ⎤⎦=3 2 log⎡⎣ _b a + −1 log_b c −log_b d ⎤⎦ Distribute.

( ) ( ) ( ) ( ) ( ) ( )

3 2 log⎡⎣ _b a + −1 log_b c −log_b d ⎤⎦=6 log_b a + −3 3log_b c −3log_b d Expand

2 3

log_b a b cd

⎡⎛ ⎞ ⎤

⎢⎜ ⎟ ⎥

⎢⎝ ⎠ ⎥

⎣ ⎦

as a sum or difference of logarithms. Remove all exponents from the argument.

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Example 4

Rewriting Sums and Differences of Logarithms as Single Logarithms

(

MN

)

=^logb

( )

M +^logb

( )

N .

( ) ( ) ( )( )

(

²

)

log 5 log 5 log 5 5

= log 25

b b

b

x x x x

x

⎡ ⎤

+ + − = ⎣ + − ⎦

−

Example 5

Rewriting Sums and Differences of Logarithms as Single Logarithms

(

M N

)

=^logb

( )

M −^logb

( )

N .

( ) ⁽ ⁾

( ) ( )

2

2 9

log 9 log 3 log

3

3 3

= log

b b

b

x x x

x

x x

⎡ − ⎤

− − − = ⎢⎣ − ⎥⎦

+ −

3 x−

( )

= log_b x 3

⎡ ⎤

⎢ ⎥

⎣ ⎦

+

Example 6 Estimating Logarithms

Find a power of three that approximates the given argument. Approximate the given logarithm using an approximate logarithm with a power of three as the argument.

( ) ( )

ln 250 ≈ln 243

( )

^M^p ^{= ⋅}^p ^log^b

^{( )}

^M ^.

( ) ( ) ( )

⁵

^{( )}

ln 250 ≈ln 243 =ln 3 = ⋅5 ln 3 ≈ ⋅5 1.1≈5.5 Rewrite ^logb

(

x+⁵

)

+^logb

(

x−⁵

)

as a single logarithm.

Rewrite ^log^b

(

^x² ⁻⁹

)

⁻^log^b

⁽

^x⁻³

⁾

as a single logarithm.

Given that ^{ln 3}

( )

⁼^1.1, estimate the value of ^{ln 250}

( )

^.

(12)

Suggested Homework from Blitzer Section 4.3: #1-57 odd

Application Exercise

In optics, the transparency, T, of a substance equals the ratio of the intensity of light transmitted through the substance

( )

JT to the intensity of incident light

( )

JI where incident light is the light falling on the surface of a substance.

The opacity of a substance equals the reciprocal of transparency. The density, D, of a substance equals the common logarithm of the opacity. Find density in terms of intensity of transmitted and incident light then expand the equation to a difference of two logarithms.

(13)

ANSWERS

#1) 3 #3) ½ #5) –1 #7) 4

#9) ³/₂ #11) 4.5 #13) 1 #15) log2x3

#17) 5.5

Evaluate the following expressions.

#1 log₂8 #2 log10, 000

#3 log 4 ₁₆ #4 log7

 

¹7

#5 ^{log 0.1}

 

^#6 log4



256



#7 log₃81 #8 ^ln

 

e¹⁴

#9 log₉27 #10 log 3 ₈₁

#11 lne⁴^.⁵ #12 log81

 

¹3

#13 ln

 

¹2e ln 2

 

#14 ^10ln8_{5ln 2} 1

Write each sum, difference, or product of logarithms as a single logarithm.

#15 ^log



¹²



^log2⁽ ⁴⁾ 2

2 x x  x #16 log2

 

¹6 log (12)2

Approximate the following logarithms given that ln(3) ≈ 1.1.

#17 ln(277) #18 ln(6,600)