Logarithms: Formulas, Rules, Tricks and Solved Examples

Logarithms express the power to which a fixed base must be raised to obtain a given number. This page covers log basics, logarithm formulas, laws of logarithms, change of base, standard values and quick methods for solving logarithm questions in quantitative aptitude.

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What Are Logarithms?

For a > 0, a ≠ 1 and x > 0, logₐ x = y means aʸ = x. Thus, a logarithm gives the exponent required to produce x from the base a.

Here, a is the base, x is the argument and y is the logarithm. For example, log₂ 8 = 3 because 2³ = 8. When the base is 10, log x is commonly written without the base; when the base is e, the notation is ln x, where e ≈ 2.718.

Logarithms Formula & Tricks

Important Formulas

Definition of logarithm
logₐ x = y ⇔ aʸ = x

The logarithmic form and exponential form are equivalent. The base must be positive and not equal to 1, and the argument must be positive.

Product rule
logₐ(xy) = logₐ x + logₐ y

The logarithm of a product equals the sum of the logarithms of its positive factors.

Quotient rule
logₐ(x/y) = logₐ x − logₐ y

The logarithm of a quotient equals the difference of the logarithms of the numerator and denominator.

Power rule
logₐ(xⁿ) = n logₐ x

An exponent in the argument can be brought in front as a multiplier.

Change of base
logₐ x = log_b x / log_b a

The base can be changed to any valid base b. In particular, logₐ x = ln x / ln a.

Reciprocal and conversion rules
logₐ b = 1/log_b a; logₐ 1 = 0; logₐ a = 1

These identities follow directly from the definition of logarithms.

Quick Tricks

Convert to exponential form first

When the logarithm has a simple value, rewrite it as a power statement before calculating.

Example: log₃ 81 = x means 3ˣ = 81 = 3⁴, so x = 4.
Use prime factors for products

Break a number into powers of the base or its prime factors to evaluate the logarithm quickly.

Example: log₂ 32 = log₂(2⁵) = 5.
Apply the power rule before approximating

Separate perfect powers before using change of base or decimal values.

Example: log₅ 125^(2/3) = (2/3)log₅(5³) = (2/3) × 3 = 2.
Check the domain before solving

Every logarithm argument must be positive. This removes invalid values introduced while solving an equation.

Example: For log₂(x − 1), the condition is x − 1 > 0, so x > 1.

Logarithms Concepts

Basic logarithm values and identities

The basic values are logₐ 1 = 0 and logₐ a = 1 for every valid base a.

Since a⁰ = 1 and a¹ = a, these values follow directly from the definition. Also, logₐ(aⁿ) = n and a^(logₐ x) = x for every x > 0.

Example: log₇ 1 = 0, log₇ 7 = 1 and log₇(7⁴) = 4.

Laws of logarithms

Products become sums, quotients become differences and powers become coefficients under logarithm laws.

For positive x and y, logₐ(xy) = logₐ x + logₐ y, logₐ(x/y) = logₐ x − logₐ y and logₐ(xⁿ) = n logₐ x. These laws cannot be applied to a sum or difference inside the argument; generally, logₐ(x + y) is not equal to logₐ x + logₐ y.

Example: log₂ 12 − log₂ 3 = log₂(12/3) = log₂ 4 = 2.

Change of base and reciprocal relation

Change of base converts a logarithm into a ratio of logarithms with a convenient common base.

The formula logₐ x = ln x/ln a is useful when the required base is not directly available. The reciprocal relation logₐ b = 1/log_b a follows by applying the change-of-base formula.

Example: log₂ 8 = log 8/log 2 = 3, because 10³ is not needed; the ratio still equals 3.

Solving logarithmic equations

To solve a logarithmic equation, combine logarithms when possible, convert to exponential form and verify the domain.

For the same valid base, logₐ f(x) = logₐ g(x) implies f(x) = g(x), provided f(x) > 0 and g(x) > 0. If logₐ f(x) = k, then f(x) = aᵏ. Any candidate that makes a logarithm argument non-positive must be rejected.

Example: log₂(x − 1) = 3 gives x − 1 = 2³ = 8, so x = 9. The domain x > 1 is satisfied.

Sign and monotonicity of logarithms

For a > 1, logₐ x is positive when x > 1 and negative when 0 < x < 1; for 0 < a < 1, these signs are reversed.

If a > 1, logₐ x increases as x increases. If 0 < a < 1, logₐ x decreases as x increases. In both cases, logₐ 1 = 0 and the argument must remain positive.

Example: log₂(1/4) = −2 because 2⁻² = 1/4, while log_(1/2)(1/4) = 2 because (1/2)² = 1/4.

Logarithms Video Lessons

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13 Lessons
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Logarithms: Definition and Fundamental Laws

Understand the definition of logarithms and learn their fundamental laws, including the rules for simplifying logarithmic expressions in quantitative aptitude.

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Practice Logarithms Questions

Practise published questions related to this topic.

1If 5^(x+1) = 1 , what is the value of x?→ 2Find the value of { (25 × 4⁻⁴)/(5⁻³ × 10 × 4⁻⁸) } ÷ { (3⁻⁵ × 10⁻⁵ × 125)/(5⁻⁷ × 6⁻⁵) } .→ 3Find the value of √8+2√32-√128+2√50 .→ 4An officer checks the cost ratio: √18 ÷ √2 .Simplify.→ 5Find the value of: (1/((5-√6))) + (1/((5+√6))) - (10/((25-6)))→ 6An officer simplifies 1/(√3 + √2) . The simplified form is:→ 7Simplify: (∛4096/⁴√10000) × 5/48 × 186→ 8Evaluate: 16 × 2⁵ × 8 ÷ (1/2)³→ 9Simplify: ((6⁴ × 4⁵ × 6⁵))/((6² × 4² × 6³))→ 10Find the simplified value of the given expression. ((2³ × 8²)³ ÷ (4⁵ × 16²))/((32² ÷ 2⁴)²)→

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Quick Revision Notes

Logarithms Revision Points

Remember the definition, domain restrictions, standard laws and base conditions.

  • logₐ x = y is equivalent to aʸ = x.
  • The conditions are a > 0, a ≠ 1 and x > 0.
  • logₐ 1 = 0 and logₐ a = 1.
  • logₐ(xy) = logₐ x + logₐ y.
  • logₐ(x/y) = logₐ x − logₐ y.
  • logₐ(xⁿ) = n logₐ x.
  • logₐ x = ln x/ln a.
  • logₐ b = 1/log_b a; the bases are interchanged in the reciprocal form.

Logarithms FAQs

What is the value of log₂ 64?

log₂ 64 = 6 because 2⁶ = 64.

What is logₐ 1?

logₐ 1 = 0 for every valid base a because a⁰ = 1.

Can logₐ(x + y) be written as logₐ x + logₐ y?

No. The product rule applies to multiplication, not addition. Generally, logₐ(x + y) ≠ logₐ x + logₐ y.

How do you solve log₃(x) = 4?

Convert to exponential form: x = 3⁴ = 81.

What is the value of log₄ 8?

Using change of base, log₄ 8 = log₂ 8/log₂ 4 = 3/2.

What is the domain of log₅(2x − 6)?

The argument must be positive: 2x − 6 > 0. Therefore, x > 3.

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