Math Lesson 13.1.3 - The Properties of Logarithm

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Welcome to our Math lesson on The Properties of Logarithm, this is the third lesson of our suite of math lessons covering the topic of Definition and Properties of Logarithms, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

The Properties of Logarithm

Logarithms have a number of properties that derive from their relationship with exponents. All these properties are listed below.

1. Product property rule

This rule says:

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

In symbols, the product rule of logarithms is written as

Proof: Let log_am = x and log_an = y. Expressing these logarithms in the exponential form yields

Multiplying the above expressions yields

From the properties of exponents, we have

Therefore, we write

Expressing the last expression back in the logarithmic form yields

or

Hence, the product property of logarithms is already confirmed as true.

Example 2

Calculate the value of the following expressions:

1. log₈ 32 + log₈ 2
2. log₆ 24 + log₆ 9

Solution 2

1. In the actual form, it is impossible to find an exact value for the expression, as no integer can be an exponent of 8 to give 32, or an exponent of 8 to give 2. Therefore, we apply the product property rule, i.e.
   log₈ 32 + log₈ 2
   = log₈ (32 ∙ 2)
   = log₈ 64
   = 2
   This is true because 8² = 64.
2. In the actual form, it is impossible to find an exact value for the expression, as no integer can be an exponent of 6 to give 24, or an exponent of 6 to give 9. Therefore, we apply the product property rule, i.e.
   log₆ 24 + log₆ 9
   = log₆ (24 ∙ 9)
   = log₆ 216
   = 3
   This is true because 6³ = 216.

2. Quotient property rule

"The logarithm of a quotient is equal to the difference of the individual logarithms."

In symbols, the quotient rule of logarithms is written as

Proof: Let log_a m = x and log_a n = y. Expressing these logarithms in the exponential form yields

Dividing the above expressions by each other yields

From the properties of exponents, we have

Therefore, we write

Expressing the last expression back in the logarithmic form yields

or

Hence, the quotient property of logarithms is already confirmed as true.

Example 3

Calculate the value of the following expressions:

1. log₄ 80 - log₄ 5
2. log₅ 2500 - log₅ 20

Solution 3

1. It is impossible to obtain an exact value when trying to find the individual logarithms in the actual form. Therefore, in this case, we apply the quotient property rule of logarithms
   log_a (m/n) = log_a m - log_a n
   Thus,
   log₄ 80 - log₄ 5 = log₄ 80/5
   = log₄ 16
   = 2
   This result is correct because 4² = 16.
2. It is impossible to obtain an exact value when trying to find the individual logarithms in the actual form. Therefore, in this case, we apply the quotient property rule of logarithms
   log_a (m/n) = log_a m - log_a n
   Thus,
   log₅ 2500 - log₅ 20 = log₅ 2500/20
   = log₅ 125
   = 3
   This result is correct because 5³ = 125.

3. Power property rule

"The logarithm of a number raised in a given power is equal to the product of that power and the logarithm of the number itself."

In symbols, the power rule of logarithms is written as

Proof: We have

Hence, the power property of logarithms is already confirmed as true.

Example 4

Calculate the value of the following expressions:

1. log₃ 9⁴ + log₂ 32 - 3 log 1000
2. 2 log₆ 125 - log₆ 150 - log₆ 625

Solution 4

1. Applying the power property rule of logarithm for all terms of the given expression yields
   log₃ 9⁴ + log₂ 32 - 3 log 1000
   = 4 log₃ 9 + log₂ 2⁵ - 3 log 10³
   = 4 log₃ 3² + log₂ 2⁵ - 3 log 10³
   = 2 ∙ 4 log₃ 3 + 5 log₂ 2 - 3 ∙ 3 log 10
   = 8 log₃ 3 + 5 log₂ 2 - 9 log 10
   = 8 ∙ 1 + 5 ∙ 1 - 9 ∙ 1
   = 8 + 5 - 9
   = 4
2. Applying the power property rule of logarithm for all terms of the given expression (also combined with the other two properties explained earlier) yields
   2 log₆ 125 - log₆ 150 - log₆ 625
   = 2 log₆ 5³ - log₆ (6 ∙ 25) - log₆ 5⁴
   = 3 ∙ 2 log₆ 5 - (log₆ 6 + log₆ 25 ) - log₆ 5⁴
   = 6 log₆ 5 - log₆ 6 - log₆ 5² - log₆ 5⁴
   = 6 log₆ 5 - log₆ 6 - 2 log₆ 5 - 4 log₆ 5
   = 6 log₆ 5 - 2 log₆ 5 - 4 log₆ 5 - log₆ 6
   = 0 - 1
   = -1

4. Log of the same number as base property rule

In both the above examples, we have anticipated the following derived property of logarithm

This property is called the "log of the same number as base" rule.

This property is true because a¹ = a.

5. Change of base property rule

Sometimes, it is more suitable to change the base of a logarithm to make the operations easier. The change of base rule says:

It is possible to divide the logarithm of the old argument by the logarithm of the old base by expressing these new logarithms at the same new base; the result obtained does not change.

In symbols, the change of base rule of logarithms is written as

Proof: Let's express log_ab = x, log_c b = y and log_c a = z. Writing them in the exponential form yields

Combining the first two expressions yields

Substituting in the above equation a = c^z obtained from the third expression yields

Since the bases are the same, the exponents must be equal as well. Thus, we have

or

Substituting back the old variables in the above equation yields

Hence, the change of base property of logarithms is already confirmed as true.

Example 5

Find the result of the following operations.

1. log₄ 8 + log₃ 27 - log₂ 32 - 13 log 100
2. log₂ 36 - 2 log₂ 6 - log₇ 196 + 2 log₇ 14

Solution 5

1. Applying the change of base rule of logarithms to express all individual logarithms to the appropriate base yields
   log₄ 8 + log₃ 27 - log₂ 32 - 13 log 100
   = log₂ 8/log₂ 4 + log₃ 27/log₃ 3 - log₂ 32/log₂ 2 - 13 log 100
   = log₂ 2³/log₂ 2² + log₃ 3³/log₃ 3 - log₂ 2⁵/log₂ 2 - 13 log 10²
   3 log₂ 2/2 log₂ 2 + 3 log₃ 3/log₃ 3 - 5 log₂ 2/log₂ 2 - 2 ∙ 13 log 10
   = 3 ∙ 1/2 ∙ 1 + 3 ∙ 1/1 - 5 ∙ 1/1 - 2 ∙ 13 ∙ 1
   = 3/2 + 3 - 5 - 26
   = 3/2 - 28
   = 3/2 - 56/2
   = -53/2
2. Again, applying the change of base rule of logarithms to express all individual logarithms to the appropriate base yields
   log₂ 36 - 2 log₂ 6 - log₇ 196 + 2 log₇ 14
   = log₂ 6² - 2 log₂ 6 - log₇ (49 ∙ 4) + 2 log₇ (7 ∙ 2)
   = 2 log₂ 6 - 2 log₂ 6 - (log₇ 49 + log₇ 4 ) + (2 log₇ 7 + 2 log₇ 2 )
   = log₇ 7² - log₇ 2² + 2 log₇ 7 + 2 log₇ 2
   = 2 log₇ 7 - 2 log₇ 2 + 2 log₇ 7 + 2 log₇ 2
   = (2 log₇ 7 + 2 log₇ 7 ) - (2 log₇ 2-2 log₇ 2 )
   = 2 ∙ 1 + 2 ∙ 1
   = 4

6. Equality property rule

This rule says that if two equal numbers written in the logarithmic form have the same base, then their arguments are also equal.

In symbols, we write this property as

Proof: Let log_ab = x and log_ac = y. Then, we have

Since x = y, then

This means that b = c. This means the equality property of logarithm is true.

7. Number raised to log property rule

Sometimes, the exponent of a number may contain a logarithm instead of an ordinary number. The number raised to log property rule says that

If a number is raised to a logarithmic power, where the base of logarithm is the same as the base of the given number, then the value of the expression is equal to the value of the logarithm argument.

In symbols, we write this property as

Proof: Taking the base a logarithm of both sides (we can do this since both sides are equal) yields

From the power property rule of logarithms we send the exponent of the argument on the left side before the expression. In this way, we obtain

Given that log_aa = 1, we obtain

Since both sides are identical, the number raised to log property is already proven.

Example 6

Do the operations until you get the simplest form of the following expressions.

1. 5^{log₅ 12} - 3^{log₃ 81}/23 ∙ 10^{log 3}
2. (1/8)^{log₂ 8} - 5 ∙ 3^{log₃ 81}/4^{log₄ 17}

Solution 6

1. Applying the number raised to log property of logarithms yields
   5^{log₅ 12} - 3^{log₃ 81} /23 ∙ 10^{log 3} = 12 - 81/23 ∙ 3= -69/69= -1
2. (1/8)^{log₂ 8} - 5 ∙ 3^{log₃ 81}/4^{log₄ 17}
   = (1/2³ )^{log₂ 8} - 5 ∙ 81/17
   = 2^{-3 ∙ log₂ 23} - 405/17
   = 2^{-3 ∙ 3 ∙ log₂ 2} - 405/17
   = 2^{-9 ∙ 1} - 405/17
   = 2^-9 - 405/17
   = 1/2⁹ - 405/17
   = 1/512 - 405/17
   = 1 - 521 ∙ 405/512/17/1
   = 1 - 521 ∙ 405/512 ∙ 1/17
   = 1 - 521 ∙ 405/512 ∙ 17
   = 1 - 211,005/8,704
   = 211,004/8,704
   = 52,751/2,176

8. Exponential base of logarithm property rule

Sometimes, the base of logarithm is written in the exponential form. In these cases, the following property is applied:

When the base of a logarithm is expressed in the exponential form, its exponent can be written in the denominator of the fraction preceding the logarithm.

In symbols, we write this property as follows

Proof: This property is true because

Example 7

Calculate the value of the following expressions:

1. log₄ 8 + log₈ 16 - log₁₆ 64
2. 2 log₂₇ 9 - 5 log₉ 81 + 4 log₈₁ 27

Solution 7

1. Let's write all terms at base (and argument) 2 for easier calculations. Thus,
   log₄ 8 + log₈ 16 - log₁₆ 64
   = log_2² 2³ + log_2³ 2⁴ - log_2⁴ 2⁶
   Applying the exponential base of logarithm property rule yields
   log_2² 2³ + log_2³ 2⁴ - log_2⁴ 2⁶
   = 3/2 log₂ 2 + 4/3 log₂ 2 - 6/4 log₂ 2
   = 3/2 ∙ 1 + 4/3 ∙ 1 - 6/4 ∙ 1
   = 3/2 + 4/3 - 6/4
   = 3/2 + 4/3 - 3/2
   = 4/3
2. Let's write all terms at base (and argument) 3 for easier calculations. Thus,
   2 log₂₇ 9 - 5 log₉ 81 + 4 log₈₁ 27
   = 2 log_3³ 3³ -5 log_3² 3⁴ + 4 log_3⁴ 3³
   Applying the exponential base of logarithm property rule yields
   = 2/3 ∙ 2 log₃ 3 - 4/2 ∙ 5 log₃ 3 + 3/4 ∙ 4 log₃ 3
   = 4/3 ∙ 1 - 20/2 ∙ 1 + 12/4 ∙ 1
   = 4/3 - 10 + 3
   = 4/3 - 7
   = 4/3 - 21/3
   = -17/3

9. Rational base or argument property rule

This property says:

If the base or argument of a logarithm (or both) are expressed as fractions, we can write them as rational powers and then continue using any of the previous properties of logarithm.

In symbols, we write it as

Proof: This property is true because

and

For example,

Example 8

Calculate the value of the following expressions:

1. log_1/3⁡9 + 2 log 1/10 - 3 log_1/2⁡1/4
2. log₅ 1/25 - 3 log_1/4⁡8^1/2 + 4 log_1/3 1/27

Solution 8

1. Applying the rational base or argument property of logarithm yields
   log_1/3 9 + 2 log 1/10 - 3 log_1/2 1/4
   = log_{3 ^{- 1}} 3² + 2 log 10^-1 - 3 log_2^-1 2^-2
   = 2/-1 log₃ 3 + (-1) ∙ 2 ∙ log 10 - (-2)/(-1) log₂ 2
   = -2 ∙ 1 - 2 ∙ 1 - 2 ∙ 1
   = -2 - 2 - 2
   = -6
2. Again, applying the rational base or argument property of logarithm yields
   log₅ 1/25 - 3 log_1/4 8^1/2 + 4 log_1/3⁡1/27
   = log₅ 5^-2 - 3 log_2^-2 (2³)^1/2 + 4 log_3^-1 3^-3
   = log₅ 5^-2 - 3 log_2^-2⁡2^3/2 + 4 log_3^-1 3^-3
   = -2 log₅ 5 - 3 ∙ 3/2/-2 log₂ 2 + 4 ∙ -3/-1 log₃ 3
   = -2 log₅ 5 - 3 ∙ 3/2 ∙ -1/2 log₂ 2 + 4 ∙ -3/-1 log₃ 3
   = -2 ∙ 1 + 9/4 ∙ 1 + 12/1 ∙ 1
   = -2 + 9/4 + 12
   = 10 + 9/4
   = 40/4 + 9/4
   = 49/4

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