A logarithm answers a specific question: to what power must a base be raised to produce a given number?
Write log_b(N) = x and you are saying b^x = N. That is the complete definition. Nothing more, nothing less.
Here is the analogy that makes it stick. Multiplication and division are inverse operations — you can undo one with the other. The same relationship exists between exponentiation and logarithms. If exponentiation says "start at base b, apply power x, get N," then the logarithm reverses the journey: "start at the result N, tell me what power of b produced it."
So 10^3 = 1000 ↔ log₁₀(1000) = 3. You already knew the first form. The second form is simply the same fact, read from the other direction.
Two bases you will actually encounter in CDS:
log without any base specified, it always means base 10 in Indian competitive exams. This is your primary tool.e ≈ 2.718. You will not be asked to compute ln values from scratch in CDS, but you must know it exists and that ln(e) = 1.Domain restriction — this one gets people: log_b(N) is only defined when N > 0 and b > 0, b ≠ 1. You cannot take the log of a negative number or zero. This fact appears disguised in CDS questions as "for what values of x is this expression defined?" — always check domain first.
Antilog is the reverse lookup: antilog₁₀(x) = 10^x. If log₁₀(N) = 2.3, then N = antilog(2.3) = 10^2.3.
The real power of logarithms in an exam hall is not computing exact values — it is converting multiplication into addition and division into subtraction. That transformation is what all the PYQs test.
These four laws are the only machinery you need. Know them cold — not just as formulas, but as transformations you can apply in both directions.
Law 1 — Product Rule:
log_b(M · N) = log_b(M) + log_b(N)
Direction matters: you can split a log of a product into a sum, or merge a sum of logs into a log of a product. CDS questions almost always require the merge direction.
Law 2 — Quotient Rule:
log_b(M/N) = log_b(M) − log_b(N)
Law 3 — Power Rule:
log_b(M^k) = k · log_b(M)
This is your tool for pulling coefficients out front, or pushing them back in as exponents. 4 log₁₀(25) = log₁₀(25⁴) — that move alone solves half of CDS log questions.
Law 4 — Change of Base:
log_b(N) = log_a(N) / log_a(b)
In CDS, this appears in the specific form: log_b(N) = 1 / log_N(b). Commit this one. It is cleaner for exam use.
| Identity | Value |
|---|---|
| log_b(b) | 1 |
| log_b(1) | 0 |
| log_b(b^k) | k |
| b^(log_b(N)) | N |
| log_b(N) · log_N(b) | 1 |
The last one follows directly from Change of Base but appears often enough in MCQs that you should pattern-match it instantly.
Look — virtually every CDS logarithm question where you are given values like log(80), log(45), log(216) and asked to find log(384) is secretly a prime factorisation problem. The logs themselves are just notation. The real work is at the level of powers of 2, 3, and 5.
Here is the operating procedure:
log 2, log 3, log 5 using the Product and Power rules.Example (abbreviated): If log(80) = p, that means log(2⁴ · 5) = 4 log 2 + log 5 = p. Store that equation. Every given value is an equation in log 2, log 3, log 5. The target is another such equation. Combine them.
The shortcut version of the same idea: rather than solving the system, directly combine the given log expressions using arithmetic until the argument inside equals the target number. This is usually faster and is exactly what the official explanations for the PYQs below do.
log₁₀(N) for a positive N is split as:
log₁₀(N) = characteristic + mantissa
For a number with n digits before the decimal, characteristic = n − 1. For a decimal 0.00xyz... with k leading zeros after the decimal point, characteristic = −(k+1).
CDS rarely asks direct characteristic/mantissa questions, but they appear in "how many digits does 2^50 have?" type problems, where you compute 50 · log₁₀(2) ≈ 50 × 0.301 = 15.05, so 2^50 has 16 digits (characteristic 15, plus 1).
The method is always the same:
argument > 0).Step 4 is where half the marks are lost. If checking x = 0 makes an argument negative, that root is extraneous — discard it.
When you see a sum of logs to be simplified, immediately write all of them under a single log as a product: log A + log B − log C = log(A·B/C). Then factorise the inside. This avoids writing intermediate steps.
Worked micro-example: log(2000) + log(400) + 4·log(25) + 5·log(20) becomes log(2000 × 400 × 25⁴ × 20⁵) in one write. Standard method of expanding each term: 6 steps, ~90 seconds. This merge-first approach: 1 write, then prime factorisation: ~30 seconds.
Before touching any log law, write the prime factorisation of every number in the problem. Numbers like 80, 45, 216, 384 look complicated but are just small primes raised to small powers. Once factorised, the log problem becomes a system of linear equations in log 2, log 3, log 5 — which you solve by inspection.
Micro-example: 384 = 2⁷ · 3. If you see 2⁷ · 3 in your factorisations of 80 × 216 / 45, you know the answer without solving any system. Counting prime factors: 4 steps. Solving 3-variable system: 10+ steps.
When a logarithmic equation looks messy — especially when powers of x appear both inside and outside logs — test the answer choices from the easiest value first (x = 0, then x = 1). The CDS format gives you 4 options; elimination is faster than full algebra.
Micro-example: log₁₀(100001 − 4^x)/(5 − x) = 1. Plug x = 0: LHS = log(100000)/5 = 5/5 = 1 ✓. Done in 15 seconds. Algebraic rearrangement from scratch: 2+ minutes with risk of error.
Whenever you see log_a(b) · log_b(a) or log_b(N) + log_N(b) type structures, deploy the identity log_b(N) · log_N(b) = 1 immediately. These appear in simplification questions that look like multi-step algebra but collapse in one line.
Micro-example: log₂(8) · log₈(2) = 3 × (1/3) = 1. Or recognise it as log_b(N) · log_N(b) = 1 directly. Standard expansion: 4 steps. Reciprocal identity: 0 steps.
The two values log₁₀(2) = 0.301 and log₁₀(3) = 0.477 generate any common log you need: log 5 = log(10/2) = 1 − 0.301 = 0.699, log 6 = log 2 + log 3 = 0.778, log 7 ≈ 0.845, log 8 = 3 × 0.301 = 0.903, log 9 = 2 × 0.477 = 0.954. Memorise just two numbers and derive the rest. Number-of-digits problems resolve in under 20 seconds once you have n · log(base).
In the exam hall, classify the question in the first 5 seconds:
Type A — Simplify a sum/difference of logs to a number:
Merge all terms into a single log using Product/Quotient rules and the Power rule. Then prime factorise the argument. If it equals 10^k, the answer is k. Time budget: 60 seconds.
Type B — Express one log in terms of given log values:
Prime factorise all numbers. Write each given value as an equation in log 2, log 3, log 5. Express the target the same way. Combine the given equations. Or: directly try arithmetic combinations of the given values (p ± q ± r) and see which one gives the target argument when converted back. Time budget: 90 seconds.
Type C — Solve a logarithmic equation for x:
Try the answer choices from simplest (x = 0, x = 1) first. If none work immediately, convert to exponential form and solve algebraically. Always verify domain at the end. Time budget: 45-60 seconds with substitution, 2+ minutes algebraically.
Type D — Change of base / identity simplification: Spot the reciprocal identity or chain-of-logs pattern. These are 10-second questions if you know the identities cold.
If the question involves log of numbers you cannot factorise cleanly, it is likely a plug-and-verify question — go straight to the options.
Why this question: The premier example of Type B — three given log values, one target. Tests whether you think in prime factorisation.
Solving path: Factorise everything: 80 = 2⁴·5, 45 = 3²·5, 216 = 2³·3³, 384 = 2⁷·3. Now compute p − q + r = log(80) − log(45) + log(216) = log(80 × 216 / 45). Numerator: 80 × 216 = 17280. Divide: 17280 / 45 = 384. So log(384) = p − q + r. The answer is option B. The entire question is arithmetic on prime factors — zero log theory required once you see the structure.
Why this question: Classic Type A. A sum of four log terms that looks intimidating but collapses to 10^18 through prime factorisation.
Solving path: Convert all coefficient-out terms using the Power rule: 4 log(25) = log(25⁴) = log(5⁸), 5 log(20) = log(20⁵) = log(2¹⁰ · 5⁵). Now merge: log(2000 × 400 × 5⁸ × 2¹⁰ · 5⁵). Factorise 2000 = 2·10³ = 2·2³·5³ = 2⁴·5³ — wait, cleaner: 2000 = 2⁴·5³·... Let us just track powers of 2 and 5 separately. 2000 = 2⁴·5³, 400 = 2⁴·5². Running total: 2^(4+4+10) · 5^(3+2+8+5) = 2^18 · 5^18 × 10^5. Actually follow the official explanation: 2000 × 400 × 5⁸ × 2¹⁰·5⁵ = (2·10³)(2²·10²)(5⁸)(2¹⁰·5⁵) = 2¹³ · 5¹³ · 10⁵ = 10¹³ · 10⁵ = 10¹⁸. Answer: 18.
Why this question: The Type C case — an equation with x inside and outside the log simultaneously. Direct substitution (x = 0) is the intended fast path.
Solving path: The equation is log₁₀(100001 − 4^x)/(5 − x) = 1. This means log₁₀(100001 − 4^x) = 5 − x, so 100001 − 4^x = 10^(5−x). Try x = 0: LHS = 100001 − 1 = 100000. RHS = 10^5 = 100000. Match. Answer: 0. You do not need to check whether x = 1, x = 10, or x = 100 also work — CDS is single-correct, and x = 0 satisfies the domain (argument = 100000 > 0, denominator 5 − 0 = 5 ≠ 0). Total time: 15 seconds.
Forgetting the domain check in equation questions. After solving algebraically, always substitute back and verify that every log argument is strictly positive. A common trap: an equation has two algebraic roots, but only one satisfies the domain. CDS questions are sometimes built around this.
Misapplying the Power rule direction. log(M^k) = k · log(M) — the exponent comes out front as a multiplier. Students sometimes write log(k · M) instead of k · log(M). These are entirely different things. log(k · M) = log k + log M by the Product rule.
Treating log(M + N) as log M + log N. The Product rule applies to multiplication inside the log, not addition. log(80 + 20) ≠ log 80 + log 20. This mistake is easy to make under time pressure when you see a sum and pattern-match to the Product rule.
Using base 10 where the problem implies a different base. If a question says log₂(x) = 3, then x = 2³ = 8, not x = 10³. Always check the base before converting.
Ignoring that log_b(1) = 0 when simplifying. Expressions like log₂(1) + log₂(8) simplify immediately to 0 + 3 = 3. Students sometimes get stuck computing log₂(1) by working out 2^x = 1 from scratch instead of recalling the identity.
Confusing log and ln in a question. In Indian competitive exam notation, unsubscripted log always means log₁₀. ln is reserved for the natural logarithm. If you mix them up in a numerical question, your answer will be off by a factor of log₁₀(e) ≈ 0.434.