Logarithms

Logarithms look scary at first, but they’re just another way of writing exponents. If we understand that log is the inverse of exponentiation, everything clicks.

In simple language: if 2³ = 8, then log₂(8) = 3. We’re asking “what power do we raise 2 to, to get 8?” The answer is 3. That’s literally all a log is — the exponent finder.

The Log Laws

These are the tools that make simplification problems solvable in seconds. Every log problem in exams boils down to applying these laws.

Think of the product rule like this: multiplication inside the log becomes addition outside. Division becomes subtraction. And exponents come out front as multipliers. Once these three rules are second nature, we can tear through any log problem.

Values to Memorize

These come up constantly. Memorizing them saves precious seconds.

The big trick: We only need to memorize log 2, log 3, and log 7. Everything else can be derived:

• log 4 = 2 × log 2
• log 5 = log(10/2) = 1 - log 2
• log 6 = log 2 + log 3
• log 8 = 3 × log 2
• log 9 = 2 × log 3

So really, just remember: log 2 ≈ 0.301, log 3 ≈ 0.477, log 7 ≈ 0.845.

Worked Examples

Example 1: Basic simplification

Find: log₂(32)

32 = 2⁵, so log₂(32) = 5

Always try to express the number as a power of the base first. That’s the fastest approach.

Example 2: Using log laws

Simplify: log 72

Break 72 into prime factors: 72 = 8 × 9 = 2³ × 3²

log 72 = log(2³ × 3²) = 3 log 2 + 2 log 3 = 3(0.301) + 2(0.477) = 0.903 + 0.954 = 1.857

Example 3: Change of base

Find: log₈(32)

Using change of base: log₈(32) = log(32)/log(8) = log(2⁵)/log(2³) = 5 log 2 / 3 log 2 = 5/3

Shortcut: When both the base and number are powers of the same thing, just use the exponents. 32 = 2⁵ and 8 = 2³, so log₈(32) = 5/3. Done!

Example 4: The flip rule in action

If log₃(5) = x, find log₅(27) in terms of x.

log₅(27) = log₅(3³) = 3 × log₅(3)

Now, log₅(3) = 1/log₃(5) = 1/x (using the flip rule)

So log₅(27) = 3/x

Example 5: Finding number of digits

How many digits are in 2⁶⁴?

The number of digits in N = floor(log₁₀(N)) + 1.

log₁₀(2⁶⁴) = 64 × log₁₀(2) = 64 × 0.3010 = 19.264

Number of digits = floor(19.264) + 1 = 19 + 1 = 20 digits

This is a classic exam question type. The formula works because log tells us the “order of magnitude” of a number.

Common Traps and Gotchas

1. log(A + B) ≠ log A + log B — There is NO rule for log of a sum! This is the most common mistake. The product rule only works with multiplication inside the log.

2. log(0) is undefined — We can’t take log of zero or negative numbers (in real numbers).

3. The base must be positive and not 1 — log₁(anything) is undefined.

4. log(A) - log(B) ≠ log(A - B) — It equals log(A/B).

5. Don’t confuse log and ln — “log” usually means base 10 (common log), “ln” means base e ≈ 2.718 (natural log). In aptitude exams, log almost always means base 10 unless stated otherwise.

Natural Logarithm (ln) — Quick Basics

The natural log uses base e ≈ 2.718 (Euler’s number). It shows up in continuous growth/decay problems.

Key facts:

• ln(e) = 1
• ln(1) = 0
• ln(eˣ) = x
• e^(ln x) = x
• All the same log laws apply (product, quotient, power)

For aptitude exams, natural log questions are rare, but knowing these basics helps if one pops up.

Common Exam Variations

• Simplify log expressions using log laws.
• Find the value of logₐ(b) when both a and b are given as powers.
• “If log x = 1.732, find x” — this means x = 10^1.732.
• Number of digits in large powers.
• Solve equations with logs: “if log(x + 2) + log(x - 1) = log 10” → log[(x+2)(x-1)] = 1 → (x+2)(x-1) = 10.
• Characteristic and mantissa (the integer and decimal parts of a log).

Characteristic and Mantissa

For log₁₀(N):

• Characteristic = the integer part (tells us how many digits minus 1)
• Mantissa = the decimal part (always positive, 0 ≤ mantissa < 1)

For numbers > 1: characteristic = (number of digits - 1). For numbers between 0 and 1: characteristic is negative. E.g., log(0.005) has characteristic -3.

Practice Problems

Problem 1: Simplify: log₅(125) + log₅(1/25).

Problem 2: If log 2 = 0.301 and log 3 = 0.477, find the number of digits in 6²⁰.

Problem 3: Solve for x: log₃(x) + log₃(x - 8) = 2.

Answers

Problem 1: log₅(125) = log₅(5³) = 3. log₅(1/25) = log₅(5⁻²) = -2. Answer: 3 + (-2) = 1.

Problem 2: log(6²⁰) = 20 × log 6 = 20 × (log 2 + log 3) = 20 × (0.301 + 0.477) = 20 × 0.778 = 15.56. Number of digits = 15 + 1 = 16 digits.

Problem 3: log₃(x) + log₃(x - 8) = 2 → log₃[x(x - 8)] = 2 → x(x - 8) = 3² = 9. So x² - 8x - 9 = 0 → (x - 9)(x + 1) = 0. x = 9 or x = -1. Since log requires positive arguments, x must be > 8. x = 9.