📘 Algebra · Chapter MN07🎯 NDA Level : High Priority
Logarithms are the inverse of exponentiation and appear throughout NDA Mathematics — in solving exponential equations, simplifying expressions, and as a tool in other chapters. The chapter is mostly rule-based: once the definition and five core properties are clear, most questions are straightforward applications of those rules.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Converting between logarithmic and exponential form (definition);
(2) Simplifying expressions using product, quotient, and power rules;
(3) Change of base formula to evaluate or compare logs in different bases;
(4) Finding characteristic and mantissa of log₁₀ of given numbers;
(5) Solving equations of the form log_a(x) = b or a^x = b;
(6) Common values: log 2, log 3, log 5 used in calculation;
(7) Solving simultaneous equations involving logarithms.
Topics at a Glance
① Definition
log_a x = y ↔ aʸ = x, base conditions
② Core Properties
Product, quotient, power rules
③ Change of Base
log_a b = log c / log a
④ Common Values
log 2, log 3, log 5, log 7
⑤ Characteristic & Mantissa
log tables, integer + decimal parts
⑥ Equations
Exponential & logarithmic equations
1. Definition of Logarithm
1.1
The Fundamental Definition — Log as Inverse of Exponent
Everything in this chapter flows from this one equivalence
The logarithm answers the question: "To what power must the base a be raised to get x?"
loga x = y
⟺
ay = x
Read: "log base a of x equals y" means "a to the power y equals x"
⚡ Conditions on Base and Argument
For log_a x to be defined:
Base a: a > 0, a ≠ 1 (base must be positive and not 1)
Argument x: x > 0 (logarithm of a negative number is undefined)
The base a ≠ 1 because 1 raised to any power is always 1 → ambiguous.
The argument x > 0 because aʸ is always positive for any real y.
Examples of conversions:
log₂ 8 = 3 because 2³ = 8
log₁₀ 100 = 2 because 10² = 100
log₃ (1/9) = −2 because 3⁻² = 1/9
log₅ 1 = 0 because 5⁰ = 1
log_a a = 1 because a¹ = a (always)
log_a 1 = 0 because a⁰ = 1 (always)
The two results log_a(a) = 1 and log_a(1) = 0 hold for ANY valid base a. Memorise these — they appear directly in NDA simplification questions.
Common (Base 10) Logarithm
Written as log x (base 10 is implied when no base shown)
Also called common logarithm or Briggsian logarithm
log 1 = 0, log 10 = 1, log 100 = 2, log 1000 = 3
log 0.1 = −1, log 0.01 = −2
Used in log tables for calculations
Characteristic and mantissa apply to base 10
Natural (Base e) Logarithm
Written as ln x (e ≈ 2.71828)
Also called natural logarithm or Napierian logarithm
ln 1 = 0, ln e = 1, ln e² = 2
Used in calculus, growth/decay problems
ln x = log x / log e = log x / 0.4343
Relation: ln x = 2.303 × log x
Graph of log_a x for different bases
Fig 1: Graph of log_a(x). For a > 1 (solid): increasing curve through (1,0). For 0 < a < 1 (dashed): decreasing curve through (1,0). Both have the same domain x > 0.
📋 TOPIC-WISE PYQ
Definition & Conversion — NDA-Pattern Questions
Q1. If log₂ 32 = x, then x = ?
(a) 4 (b) 5 (c) 6 (d) 16
Answer: (b) 5
log₂ 32 = x means 2ˣ = 32 = 2⁵ → x = 5.
Q2. The value of log₃ (1/27) is:
(a) 3 (b) −3 (c) 1/3 (d) −1/3
Answer: (b) −3
log₃(1/27) = log₃(3⁻³) = −3.
Q3. If log_x 64 = 2, find x.
(a) 4 (b) 8 (c) 16 (d) 32
Answer: (b) 8
log_x 64 = 2 → x² = 64 → x = √64 = 8. (x must be positive and ≠ 1.)
2. Properties of Logarithms
2.1
The Five Core Log Rules
All simplification and equation-solving uses exactly these rules
① Product Rule
log_a (MN) = log_a M + log_a N
log 6 = log 2 + log 3
② Quotient Rule
log_a (M/N) = log_a M − log_a N
log(5/2) = log 5 − log 2
③ Power Rule
log_a (Mⁿ) = n · log_a M
log 8 = log 2³ = 3 log 2
④ Change of Base
log_a b = log_c b / log_c a
log₂ 8 = log 8 / log 2 = 3
⑤ Identity Rules
log_a a = 1 | log_a 1 = 0
log₅ 5 = 1; log₁₀ 1 = 0
⑥ Reciprocal Rule
log_a b = 1 / log_b a
log₂ 8 = 1 / log₈ 2 = 3
⚡ Additional Useful Results
log_a (aⁿ) = n (power and base cancel)
a^(log_a x) = x (inverse property — very important)
log_a b × log_b c = log_a c (chain rule — telescoping logs)
log_a b × log_b a = 1 (product of reciprocal logs = 1)
If log_a M = log_a N, then M = N (one-to-one property of logs)
log_a M = log_a N + log_a k → M = k·N (combine before equating)
The chain rule log_a b × log_b c = log_a c is frequently used in NDA to simplify products of logarithms with different bases.
⚠ Critical Restriction — Logarithm is NOT Linear:
• log(M + N) ≠ log M + log N (there is NO addition rule for arguments)
• log(M − N) ≠ log M − log N (quotient rule works only for M/N, not M−N)
• log(M × N) = log M + log N ← this IS correct (product rule)
This is the most common misconception tested in NDA statement-based questions.
These values are the building blocks for every numerical log problem
⚡ Change of Base Formula
log_a b = log_c b / log_c a (c is any valid base, usually 10 or e)
Most common form:
log_a b = log b / log a (switching to base 10)
log_a b = ln b / ln a (switching to base e)
Key consequence — reciprocal:
log_a b = 1 / log_b a
NDA use: evaluate log₂ 5 using base-10 table:
log₂ 5 = log 5 / log 2 = 0.6990 / 0.3010 ≈ 2.322
The change of base formula lets you compute any log using a base-10 table. In NDA, this is used to compare logs with different bases or convert them into a common form for equations.
Standard Log Values (Base 10) — Memorise These
Number
Logarithm (base 10)
Derived From
NDA Use
log 1
0
10⁰ = 1
Directly tested
log 2
0.3010
Standard value
Most used building block
log 3
0.4771
Standard value
Second most used
log 4
0.6020
2 × log 2
= 2 log 2
log 5
0.6990
log(10/2) = 1 − log 2
Key derivation
log 6
0.7781
log 2 + log 3
Product rule
log 7
0.8451
Standard value
Occasionally given
log 8
0.9031
3 × log 2
= 3 log 2
log 9
0.9542
2 × log 3
= 2 log 3
log 10
1
10¹ = 10
Directly tested
log 12
1.0792
log(4×3) = 2log2+log3
Composite
log 50
1.6990
log(100/2) = 2−log2
Common in NDA
📌 Derive Rather Than Memorise:
You only NEED to memorise log 2 = 0.3010, log 3 = 0.4771, and log 7 = 0.8451. Everything else can be derived:
log 4 = 2×0.3010 = 0.6020 | log 5 = 1 − 0.3010 = 0.6990 | log 6 = 0.3010 + 0.4771 = 0.7781 | log 8 = 3×0.3010 = 0.9031 | log 9 = 2×0.4771 = 0.9542.
📋 TOPIC-WISE PYQ
Change of Base & Values — NDA-Pattern Questions
Q8. If log 2 = 0.3010, how many digits does 2²⁰ have?
Used with log tables — characteristic tells digit count; mantissa gives precision
Any logarithm (base 10) can be written as: log₁₀ N = characteristic + mantissa, where the characteristic is the integer part and the mantissa is the decimal part (always ≥ 0).
Decimal part. Always positive. Read from log tables.
⚡ Rules for Characteristic
For a number N ≥ 1 (positive integer or decimal ≥ 1):
Characteristic = (number of digits before decimal point) − 1
Examples:
N = 5 → 1 digit → characteristic = 0
N = 45 → 2 digits → characteristic = 1
N = 253 → 3 digits → characteristic = 2
N = 1847 → 4 digits → characteristic = 3
For a number 0 < N < 1 (between 0 and 1):
Characteristic = −(number of zeros after decimal point before first non-zero digit + 1)
Written using bar notation (negative characteristic, positive mantissa):
N = 0.5 → characteristic = 1̄ (−1) ← bar means negative
N = 0.05 → characteristic = 2̄ (−2)
N = 0.007 → characteristic = 3̄ (−3)
Full value: log 0.05 = 2̄.some_mantissa = −2 + mantissa (e.g. −1.301 = 2̄.699)
The mantissa is ALWAYS positive and is the same for all numbers with the same sequence of significant digits, regardless of the decimal point position. E.g., log 253, log 25.3, log 2.53, log 0.253 all have the same mantissa (0.403) — only the characteristic differs.
Number N
log₁₀ N
Characteristic
Mantissa
Digit rule
2530
3.403
3
0.403
4 digits → 3
253
2.403
2
0.403
3 digits → 2
25.3
1.403
1
0.403
2 digits → 1
2.53
0.403
0
0.403
1 digit → 0
0.253
1̄.403 = −0.597
−1 (written 1̄)
0.403
0 zeros → −1
0.0253
2̄.403 = −1.597
−2 (written 2̄)
0.403
1 zero → −2
Number of Digits Formula (NDA Application):
The number of digits in a positive integer N = ⌊log₁₀ N⌋ + 1.
Example: Find digits in 2¹⁰. log(2¹⁰) = 10 × 0.3010 = 3.010. Digits = ⌊3.010⌋ + 1 = 4. So 2¹⁰ = 1024 has 4 digits ✓.
Convert between forms, apply properties, isolate the unknown
⚡ Standard Equation-Solving Strategies
Type 1 — Direct log equation:
log_a x = b → x = aᵇ
Example: log₃ x = 4 → x = 3⁴ = 81
Type 2 — Exponential equation:
aˣ = b → x = log_a b = log b / log a
Example: 2ˣ = 32 → x = log₂ 32 = 5
Type 3 — Log equation with variable in argument:
log(x+2) + log(x−1) = log(x+6)
→ log[(x+2)(x−1)] = log(x+6)
→ (x+2)(x−1) = x+6
→ solve the resulting equation (check domain at end)
Type 4 — Simultaneous log equations:
Use substitution: let a = log x, b = log y, solve linear system.
Type 5 — Exponential with same base:
aᶠ⁽ˣ⁾ = aᵍ⁽ˣ⁾ → f(x) = g(x) (exponents must be equal)
Always check the domain at the end. Arguments of logarithms must be positive. A valid algebraic solution may fail the domain check (e.g., x = −5 giving log(−5+2) = log(−3) — undefined).
Worked Example — Log Equation with Variable
Solve: log(x + 1) + log(x − 1) = log 8
log[(x+1)(x−1)] = log 8 (product rule left side)
x² − 1 = 8 → x² = 9 → x = ±3.
Domain check: x + 1 > 0 AND x − 1 > 0 → x > 1. So x = −3 is rejected.
Answer: x = 3.
Worked Example — Exponential Equation
Solve: 3^(2x−1) = 27
27 = 3³. Same base → exponents equal: 2x − 1 = 3 → 2x = 4 → x = 2.
Worked Example — Exponential Equation (Different Bases)
Solve: 5ˣ = 12
Taking log both sides: x·log 5 = log 12.
x = log12/log5 = (log4+log3)/log5 = (2log2+log3)/(1−log2).
Using values: x = (0.6020+0.4771)/0.6990 = 1.0791/0.6990 ≈ 1.544.
Convert all to base 2 using change of base:
log₂ x + log₂ x / log₂ 4 + log₂ x / log₂ 8
= log₂ x + log₂ x / 2 + log₂ x / 3
= log₂ x (1 + 1/2 + 1/3) = log₂ x · (11/6) = 11/6.
So log₂ x = 1 → x = 2. Key: Convert all logs to the same base before adding. The LCM of 1, 2, 3 gives 11/6 as the combined coefficient.
🧩 T5. If log(x + y)/5 = (log x + log y)/2, show that x/y + y/x = 23.
log(x+y)/5 = (log x + log y)/2 = log(xy)/2 = log√(xy).
Equating arguments: (x+y)/5 = √(xy).
(x+y)² = 25xy → x²+2xy+y² = 25xy → x²+y² = 23xy.
Divide by xy: x/y + y/x = 23. Proved. The key step is equating arguments once both sides are a single log. Then algebraic manipulation gives the result.
🧩 T6. If 2^x = 3^y = 12^z, show that z = xy/(x+2y).
Let 2^x = 3^y = 12^z = k (say).
Then: 2 = k^(1/x), 3 = k^(1/y), 12 = k^(1/z).
But 12 = 4 × 3 = 2² × 3 = k^(2/x) × k^(1/y) = k^(2/x + 1/y).
Also 12 = k^(1/z).
So 1/z = 2/x + 1/y = (2y + x)/(xy).
Therefore z = xy/(x+2y). Proved. Classic NDA proof technique: set all equal to k, express each base as a power of k, then use the factored form of the third number.
📋 Master Formula Sheet — MN07 Logarithms
All critical formulae for rapid pre-exam revision.
This material is for personal NDA exam preparation only.
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