banner

Logarithm Rules – Explanation & Examples

What is a logarithm? why do we study them? And what are their rules and laws?

To start with, the logarithm of a number ‘b’ can be defined as the power or exponent to which another number ‘a’ must be raised in order to produce the result equal to the number b.

This statement can be represented symbolically as;

log a b = n.

Similarly, we can define the logarithm of a number as the inverse of its exponents. For instance, log a b = n can be represented in exponential form as; a n = b.

Therefore, we can conclude that;

an = b ⇔ log a b = n.

Although logarithms are taught in schools for simplifying computation involving large numbers, they still have a significant role in our daily life.

Let’s see some of these applications of logarithms:

  • We use logarithms to measure acidity and alkalinity of chemical solutions.
  • Measurement of earthquake intensity is performed on the Richter scale using logarithms.
  • The level of noise is measured in dB (decibels) on a logarithmic scale.
  • Exponential processes such as decay of ratio active isotopes, growth of bacteria, spread of an epidemic in a population and cooling of a dead body are analyzed using logarithms.
  • Logarithm is used to calculate the payment period of a loan.
  • In calculus, logarithm is used to differentiate complex problems and also to determine the area under curves

Like exponents, logarithms too have rules and laws that work the same way as the rules of exponents. It is important to note that, the laws and rules of logarithms apply to logarithms of any base, however the same base must be used throughout a calculation.

We can use laws and rules of logarithms to perform the following operations:

  • Changing logarithmic functions to exponential form.
  • Addition
  • Subtraction
  • Multiplication
  • Division
  • Expanding and condensing
  • Solving logarithmic equations.

Laws of logarithms

The logarithmic expressions can be written in different ways but under certain laws called laws of logarithms. These laws can be applied on any base but during calculation the same base is used.

The four basic laws of logarithms include:

The Product Rule Law

The first law of logarithms state that the sum of two logarithms is equal to the product of the logarithms.  The first law is represented as;

⟹ log A + log B = log AB

Example:

  1. log 2 5 + log 2 4 = log 2 (5 × 4) = log 2 20
  2. log 10 6 + log 10 3 = log 10 (6 x 3) = log 10 18
  • log x + log y = log (x * y) = log xy
  1. log 4x + log x = log (4x * x) = log 4x2

The Quotient Rule Law

Subtraction of two logarithms A and B is equal to dividing the logarithms.

⟹ log A − log B = log (A/B)

Example:

  1. log 10 6 – log 10 3 = log 10 (6/3) = log 10 2
  2. log 2 4x – log 2 x = log 2 (4x/x) = log 2 4

The Power Rule Law

⟹ log A n = n log A

Example:

  1. log 10 53 = 3 log 10 5
  2. 2 log x = log x2
  • log(4x)3 = 3 log (4x)
  1. 5 ln x2 = ln x (2 *5) = ln x10

Change of Base Rule Law

⟹ log b x = (log a x) / (log a b)

Example 4:

  • log 416 = (log 16) / (log 4).

Rules of Logarithms

Logarithms are a very disciplined field of mathematics. They are always applied under certain rules and regulations.

Following rules needed to be remembered while playing with logarithms:

  • Given that an= b ⇔ log a b = n, the logarithm of the number b is only defined for positive real numbers.

⟹ a > 0 (a ≠ 1), an > 0.

  • The logarithm of a positive real number can be negative, zero or positive.

Examples

  1. 32= 9 ⇔ log 3 9 = 2
  2. 54= 625 ⇔ log 5 625 = 4
  3. 70= 1 ⇔ log 7 1 = 0
  4. 2-31/8 ⇔ log 2 (1/8) = -3
  5. 10-2= 0.01 ⇔ log 1001 = -2
  6. 26= 64 ⇔ log 2 64 = 6
  7. 3– 4= 1/34 = 1/81 ⇔ log 3 1/81 = -4
  8. 10-2= 1/100 = 0.01 ⇔ log 1001 = -2
  • Logarithmic values of a given number are different for different bases.

Examples

  1. log 9 81 ≠ log 3 81
  2. log 2 16 ≠ log 4 16
  • Logarithms to the base a 10 are refereed to as common logarithms. When a logarithm is written without a subscript base, we assume the base to be 10.

Examples

  1. log 21 = log 10
  2. log 0.05 = log 10 05
  • Logarithm to the base ‘e’ are called natural logarithms. The constant e is approximated as 2.7183. Natural logarithms are expressed as ln x which is the same as log e
  • The logarithmic value of a negative number is imaginary.
  • The logarithm of 1 to any finite non-zero base is zero.
    a0=1 ⟹ log a 1 = 0.

Example:

70 = 1 ⇔ log 7 1 = 0

  • The logarithm of any positive number to the same base is equal to 1.

a1=a ⟹ log a a=1.

Examples

  1. log 10 10 = 1
  2. log 2 2 = 1
  • Given that, x = log aM then a log a M = a

Example 1

Evaluate the following expression.

log 2 ​8 + log 2 ​4

Solution

Applying the product rule law, we get;

log 2 ​8 + log 2 ​4 = log 2 (8 x 4)

= log 2 32

Rewrite 32 in exponential form to get the value of its exponent.

32 = 25

Therefore, 5 is the correct answer

Example 2

Evaluate log 3 ​162 – log 3 2

Solution

This is a subtraction expression; therefore, we apply the quotient rule law.

log 3 ​162 – log 3 2 = log 3 (162/2)

= log 3 81

Write the argument in exponential form

81 = 3 4

Hence, the answer is 4.

Example 3

Expand the logarithmic expression below.

log 3 ​ (27x 2 y 5)

Solution

log 3 ​ (27x 2 y 5) = log 3 27 + log 3 x2 + log 3 y5

= log 3 (9) + log 3 (3) + 2log 3 x + 5log 3 y

But log 3 9 = 3

Substitute to get.

= 3 + log 3 (3) + 2log 3 x + 5log 3 y

Example 4

Calculate the value of log√2 64.

Solution

⟹ log√264 = log√2 (2)6

⟹ log√264 = 6log√2(2)

⟹ log√264 = 6log√2(√2)2

⟹ log√264= 6 * 2log√2(√2)

⟹ log√264 = 12 * 2(1)

⟹ log√264 = 12

Example 5

Solve for x if log 0.1 (0.0001) = x

Solution

⟹ log0.1(0.0001) = log0.1(0.1)4

⟹ log0.1(0.0001) = 4log0.10.1

⟹ log0.1(0.0001) = 4(1)

⟹ log0.1(0.0001) = 4

Therefore, x = 4.

Example 6

Find the value of x given, 2log x = 4log3

Solution

2logx = 4log3

Divide each side by 2.

⟹ log x = (4log3) / 2

⟹ log x = 2log3

⟹ log x = log32

⟹ log x = log9

x = 9

Example 7

Evaluate log 2 (5x + 6) = 5

Solution

Rewrite the equation in exponential form

25 = 5x + 6

Simplify.

32 = 5x + 6

Subtract both sides of the equation by 6

32 – 6 = 5x + 6 – 6

26 = 5x

x = 26/5

Example 8

Solve log x +log (x−1) = log (3x + 12)

Solution

⇒ log [x (x − 1)] = log (3x + 12)

Drop the logarithms to get;

⇒ [x (x − 1)] = (3x + 12)

Apply the distributive property to remove brackets.

⇒ x2 – x = 3x + 12

⇒ x2 – x – 3x – 12 = 0

⇒ x2 – 4x – 12 = 0

⇒ (x−6) (x+2) = 0

⇒x = − 2, x= 6

Since argument of a logarithm cannot be negative, then the correct answer is x = 6.

Example 9

Evaluate ln 32 – ln (2x) = ln 4x

Solution

ln [32/(2x)] = ln 4x

Drop the natural logs.

[32/ (2x)] = 4x

32/ (2x) = 4x.

Cross multiply.

32 = (2x)4x

32 = 8x2

Divide both sides by 8 to get;

x2 = 4

x = – 2, 2

Since, we cannot have the logarithm of a negative number, then x = 2 remain to be the correct answer.

Practice Questions

  1. Evaluate log​ 4 ​​64 + log​ 4 16
  2. log​ 3 14−2log​ 3 ​​5
  3. Evaluate 2 log35 + log3 40 – 3 log3 10
  4. Condense log 24 + log 2 5
  5. Expand log3(xy3/√z)
  6. Condense the following expression 5 ln x + 13 ln (x3+ 5) – 1/2 ln (x + 1)
  7. Simplify log a28 – log a 4 as a single logarithm
  8. Solve for the value of log 5 8 + 5 (1/1000)
  9. Solve for x in the logarithm 3log 5 2 = 2log 5 X
  10. Rewrite log12 + log 5 as a single logarithm

Previous Lesson | Main Page | Next Lesson