A logarithm is simply the power to which a base must be raised to get a number. It converts hard multiplication and division into easy addition and subtraction, which is exactly why the CDS exam loves it. This Cavalier guide builds the concept from scratch, fixes the laws of logarithms, and drills the exact patterns the OTA paper repeats.
Why logarithms matter in CDS
Logarithms appear in CDS / OTA Maths almost every cycle, usually as one or two direct questions. They are high-scoring because the answer follows from a fixed set of rules — there is no lengthy figure to draw and no ambiguous wording.
The topic also feeds into other chapters. The number of digits in a large power, the time needed for money to grow at compound interest, and many simplification shortcuts all rely on logs. Mastering this short chapter therefore pays off across the paper.
In CDS the base is almost always 10 (common log) or e (natural log). When no base is written in an exam option, read it as base 10.
Definition and the log-index bridge
If ax = N, then we write loga N = x. Here a is the base, N is the number, and x is the logarithm. The two statements say the same thing in two languages.
- Index form: 25 = 32
- Log form: log2 32 = 5
Three conditions must always hold: the base a > 0, a ≠ 1, and the number N > 0. You can never take the log of zero or of a negative number, and the base 1 is forbidden because 1 raised to any power is still 1.
loga N = x ↔ ax = N, with a > 0, a ≠ 1, N > 0.
Values you must know cold
A few results come straight from the definition and save time in every question.
- loga 1 = 0 (because a0 = 1)
- loga a = 1 (because a1 = a)
- loga am = m
- a(loga N) = N
For common logs (base 10), memorise log 2 ≈ 0.3010 and log 3 ≈ 0.4771. From these you can build many others, for example log 5 = log(10÷2) = 1 − 0.3010 = 0.6990.
Once you know log 2 and log 3, you can find log 4, log 5, log 6, log 8 and log 9 without a table. This single fact answers many CDS estimation questions.
The four laws of logarithms
These four laws are the engine of the whole chapter. They all assume the same base a.
- Product law: loga(M × N) = loga M + loga N
- Quotient law: loga(M ÷ N) = loga M − loga N
- Power law: loga(Mp) = p × loga M
- Root link: loga √M = ½ loga M (a special case of the power law)
Multiplication → addition, division → subtraction, power → multiplication. This trade is the reason logarithms were invented.
log(M + N) is not log M + log N. The product law applies only to a product inside the log, never to a sum. Mixing these up is the single biggest error CDS aspirants make.
Change of base formula
Sometimes a question mixes different bases, or a calculator/table only gives base 10. The change-of-base rule lets you rewrite any log in a base you prefer.
loga N = (logb N) ÷ (logb a) for any valid base b.
Two handy results drop straight out of this formula:
- loga b = 1 ÷ logb a (reciprocal relation)
- loga b × logb c × logc a = 1 (the chain cancels)
The reciprocal relation is a favourite CDS trap: if log2 8 = 3, then log8 2 = 1÷3 immediately.
Characteristic and mantissa
For common logarithms, every value splits into two parts. The whole-number part is the characteristic and the decimal part is the mantissa (always kept positive, between 0 and 1).
- For a number with n digits before the decimal point, the characteristic is n − 1.
- For a pure decimal with m zeros right after the point, the characteristic is −(m + 1), often written with a bar, e.g. 2̅.
The mantissa depends only on the digits, not on where the decimal point sits. So log 234, log 23.4 and log 2.34 share the same mantissa but have characteristics 2, 1 and 0.
To count the digits in a big power, find its log, take the characteristic, and add 1. Number of digits = characteristic + 1.
Worked example: counting digits
A classic CDS application is finding how many digits a large power has. We solve it with the power law and the characteristic rule.
How many digits are there in the number 250? (Take log 2 = 0.3010.)
So 250 has 16 digits. Notice we never computed the actual number — the log alone gave the answer.
Worked example: simplifying a log expression
This pattern — collapsing several logs into one value — shows up directly in the paper.
Evaluate log 2 + log 5 + log 100 (all base 10).
The expression equals 3. Spotting that 2 × 5 = 10 turns a messy sum into a one-line answer.
Always look for factors that multiply to a power of the base (like 2 × 5 = 10). It is the fastest route through most CDS log sums.
Solving logarithmic equations
Two reliable methods cover most CDS equations.
Method 1: convert to index form
If log3 x = 4, rewrite as x = 34 = 81. Done.
Method 2: make bases equal, then drop the logs
If loga M = loga N, then M = N. Bring both sides to a single log with the same base, then equate the numbers inside.
Always check the final answer against the domain. A value that makes any number inside a log zero or negative must be rejected, even if the algebra looked correct.
Previous-year style question
This question mirrors the difficulty and phrasing seen in recent CDS papers.
Q. If log10 2 = 0.3010, then the number of digits in 520 is:
(a) 13 (b) 14 (c) 15 (d) 16
Answer: (b) 14. Since 5 = 10÷2, log 5 = 1 − 0.3010 = 0.6990. Then log(520) = 20 × 0.6990 = 13.98, so the characteristic is 13 and the number of digits = 13 + 1 = 14.
If a power has base 5, convert it to 10÷2 so you only ever need log 2. This trick handles a whole family of CDS questions.
Common traps to avoid
A handful of errors cost most of the lost marks in this chapter. Train yourself to spot them.
- Splitting log(M + N) into log M + log N — never valid.
- Forgetting that the base must differ from 1 and stay positive.
- Writing the mantissa as negative; the mantissa is always positive, only the characteristic can be negative.
- Confusing loga b with logb a — they are reciprocals, not equals.
- Taking the log of a negative number or zero and reporting a value instead of "not defined".
(log M)2 is not the same as log(M2). The first squares the whole log; the second is 2 log M. Read brackets carefully.
Quick recap and revision
- loga N = x means ax = N, with a > 0, a ≠ 1, N > 0.
- Product → add, quotient → subtract, power → multiply.
- Change of base: loga N = logb N ÷ logb a; reciprocal: loga b = 1 ÷ logb a.
- Digits in a number = characteristic of its log + 1.
- Know log 2 = 0.3010 and log 3 = 0.4771; build the rest from them.
- log(M + N) ≠ log M + log N — the most punished mistake.
Revise these six lines the night before the exam and practise five mixed sums. That alone secures the logarithm marks on the CDS / OTA paper.
Frequently asked questions
What base is used in CDS logarithm questions?
Almost always base 10 (common logarithm). If no base is shown in the options, treat it as base 10. Natural log (base e) appears occasionally and is written as ln or logₑ.
Why can't the logarithm of a negative number exist?
Because no real power of a positive base ever produces a negative result. Since a > 0, a raised to any real power stays positive, so log of a negative number or zero is undefined in the real number system.
How do I find the number of digits in a large power quickly?
Take the common log of the number using the power law, read its characteristic (the whole-number part), and add 1. For example, log(2^50) = 15.05, so the number has 15 + 1 = 16 digits.
What is the difference between characteristic and mantissa?
The characteristic is the whole-number part of a common log and depends on the position of the decimal point. The mantissa is the positive decimal part and depends only on the digits, so it stays the same for 234, 23.4 and 2.34.
Which two log values should I memorise for the exam?
Memorise log 2 = 0.3010 and log 3 = 0.4771. From these you can derive log 4, 5, 6, 8, 9 and many others without a log table, which covers most CDS estimation questions.
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