Logarithm Formulas and Rules for Placement Aptitude

The logarithm log_b(a) = c asks one question: to what power must base b be raised to equal a? That definition, and three operational rules built from it, covers the full range of log problems in placement aptitude tests.

The Definition: Logarithm as Inverse Exponentiation

The two-way equivalence is the foundation:

log_b(a) = c ⟺ b^c = a

Read it as: “log base b of a equals c” means “b raised to the power c equals a.” The constraint set is strict:

• Base b: must be positive and not equal to 1
• Argument a: must be positive
• Result c: any real number

Two standard bases appear in placement tests:

Base	Name	Symbol	Typical context
10	Common logarithm	log or log_10	Aptitude tests, digit-count questions
e (approx. 2.718)	Natural logarithm	ln or log_e	Calculus, engineering, growth models

If no base is written, Indian aptitude tests assume base 10.

Three inputs produce undefined results, each appearing as a trick option in MCQ sets:

• log_b(0) is undefined. No power of any valid base equals 0.
• log_b(a) where a < 0 is undefined. No real power of a positive base gives a negative result.
• log_1(a) is undefined. Base 1 is excluded because 1^c = 1 for all c.

A common wrong answer is log(0) = 0. It is not 0. It is undefined.

Core Log Rules: Product, Quotient, and Power

Three rules handle the bulk of placement aptitude log questions.

Product Rule

log_b(x × y) = log_b(x) + log_b(y)

Log of a product equals the sum of logs. A critical distinction: log_b(x + y) does not equal log_b(x) + log_b(y). The product rule applies only when the arguments are multiplied inside the log, not added. This is one of the most-tested distractor patterns in TCS NQT and AMCAT Quantitative Ability log questions.

Quotient Rule

log_b(x / y) = log_b(x) - log_b(y)

A direct corollary: log_b(1/x) = -log_b(x). Setting y = x gives log_b(1) = 0, confirming the zero identity.

Power Rule

log_b(x^n) = n × log_b(x)

This converts an exponent into a multiplier outside the log. It is the core operation in digit-count questions (see Examples 1 and 8 below).

Reference Table

Rule	Formula
Product	log_b(xy) = log_b(x) + log_b(y)
Quotient	log_b(x/y) = log_b(x) - log_b(y)
Power	log_b(x^n) = n × log_b(x)
Identity	log_b(b) = 1
Zero	log_b(1) = 0
Reciprocal	log_a(b) = 1 / log_b(a)
Change of base	log_b(a) = log_c(a) / log_c(b)

Change of Base, Reciprocal Identity, and Special Values

Change of Base

log_b(a) = log_c(a) / log_c(b)

This converts any base to a computable one. Most aptitude problems use base 10 as the target because standard log values (log_10(2) = 0.30103, etc.) are either tabulated or given in the question. Both numerator and denominator must use the same conversion base c.

Special Values

• log_b(b) = 1 because b^1 = b
• log_b(1) = 0 because b^0 = 1
• log_b(b^n) = n directly from the power rule

Notation Ambiguity

Two expressions look similar but compute differently:

• (log_b(a))^n means compute log_b(a) first, then raise the result to the power n.
• log_b(a^n) equals n × log_b(a) by the power rule.

In standard notation, log_b a^n without parentheses means log_b(a^n), because log applies to the immediately following argument. Questions that omit parentheses test whether you apply the power rule correctly.

Characteristic and Mantissa

For base-10 logs, the integer part of the result is the characteristic; the decimal part is the mantissa. Log-table reading questions in placement papers test these definitions.

The characteristic rule for log_10(N):

• If N ≥ 1: characteristic = (number of digits in the integer part of N) minus 1
• If N < 1: characteristic = negative of (number of leading zeros after the decimal point plus 1)

Number	log_10 value	Characteristic	Mantissa
5123	approx. 3.7096	3	0.7096
3456.25	approx. 3.5385	3	0.5385
0.0045	approx. -2.347	-3	0.653

The digit-count formula follows directly: the number of digits in a positive integer N is floor(log_10(N)) + 1. This is the characteristic plus 1 for any integer.

The same floor-and-boundary logic appears in number-range aptitude puzzles. The Noddy house-number puzzle uses an analogous bounding approach to narrow a search space.

Worked Examples

Standard log values used across examples (given or from standard tables):

• log_10(2) = 0.30103
• log_10(3) = 0.47712
• log_10(5) = 0.69897 (derived: 1 - log_10(2), since log_10(10) = 1 and 10 = 2 × 5)
• log_10(7) = 0.84510

Example 1: Digit count using the power rule

• Find the number of digits in 2^256.
• Step 1: Apply the power rule: log_10(2^256) = 256 × log_10(2) = 256 × 0.30103 = 77.06368
• Step 2: Number of digits = floor(77.06368) + 1 = 77 + 1 = 78
• Answer: 78 digits

Example 2: Change of base

• Evaluate log_5(1024).
• Step 1: Convert using change of base: log_5(1024) = log_10(1024) / log_10(5)
• Step 2: 1024 = 2^10, so log_10(1024) = 10 × 0.30103 = 3.0103
• Step 3: log_10(5) = 1 - 0.30103 = 0.69897
• Step 4: 3.0103 / 0.69897 ≈ 4.3062
• Answer: approx. 4.3062

Example 3: Equation using product rule

• Solve for x: log_10(2) + log_10(6x + 1) = log_10(x + 5) + 1
• Step 1: Combine LHS using the product rule: log_10[2(6x + 1)]
• Step 2: Write 1 = log_10(10), so RHS becomes log_10[10(x + 5)]
• Step 3: Equate arguments: 2(6x + 1) = 10(x + 5) gives 12x + 2 = 10x + 50, so 2x = 48, giving x = 24
• Verify: LHS = log_10(2) + log_10(145) = log_10(290). RHS = log_10(29) + log_10(10) = log_10(290). Confirmed.
• Answer: x = 24

Example 4: Product rule (direct)

• Simplify log_2(8) + log_2(4).
• Step 1: Apply the product rule: log_2(8 × 4) = log_2(32)
• Step 2: 32 = 2^5, so log_2(32) = 5
• Answer: 5

Example 5: Quotient rule

• Evaluate log_3(81/27).
• Step 1: Apply the quotient rule: log_3(81) - log_3(27)
• Step 2: 81 = 3^4 so log_3(81) = 4; 27 = 3^3 so log_3(27) = 3
• Step 3: 4 - 3 = 1
• Verify: 81/27 = 3, and log_3(3) = 1. Confirmed.
• Answer: 1

Example 6: Power rule with fractional exponent

• Evaluate log_10(1000^(1/3)).
• Step 1: Apply the power rule: (1/3) × log_10(1000)
• Step 2: log_10(1000) = log_10(10^3) = 3
• Step 3: (1/3) × 3 = 1
• Verify: 1000^(1/3) = 10, and log_10(10) = 1. Confirmed.
• Answer: 1

Example 7: Change of base giving an integer result

• Evaluate log_8(64).
• Step 1: Change to base 2: log_8(64) = log_2(64) / log_2(8)
• Step 2: 64 = 2^6 so log_2(64) = 6; 8 = 2^3 so log_2(8) = 3
• Step 3: 6 / 3 = 2
• Verify: 8^2 = 64. Confirmed.
• Answer: 2

Example 8: Digit count with a different base

• Find the number of digits in 3^20 (use log_10(3) = 0.47712).
• Step 1: Apply the power rule: log_10(3^20) = 20 × 0.47712 = 9.5424
• Step 2: Number of digits = floor(9.5424) + 1 = 9 + 1 = 10
• Answer: 10 digits

For more placement aptitude problems combining number theory with structured reasoning, FACE Prep’s two-player coin game covers dynamic programming approaches used in aptitude rounds at analytics firms.

For practice beyond these 8 examples, IndiaBix’s Logarithms section has over 50 problems grouped by difficulty, covering the digit-count and equation-solving types worked above.

Example 1 reduced 2^256 to a 78-digit count using nothing more than the power rule and the value log_10(2) = 0.30103. The same log-base-2 arithmetic sits inside the probability and entropy calculations that language models run on every token. If applied AI is on your roadmap after placements, TinkerLLM is a ₹299 entry point for experimenting with LLM APIs and prompt engineering before committing to a longer programme.