Last Updated: April 2026
Logarithms appear consistently in IPMAT Indore and occasionally in JIPMAT and IPMAT Rohtak question papers. Aspirants treat the topic as intimidating, but the entire syllabus actually rests on five identities. Master those five, drill 25 carefully chosen problems, and the topic flips from a fear-area into a guaranteed 4–6 mark cushion in the quantitative section. This guide walks you through every IPMAT-relevant concept, the formula sheet you should print and pin above your study desk, signature question types, and a 25-question practice set with full solutions. The companion quiz at the end of this post is a 10-question diagnostic you can finish in 12 minutes.
Why Logarithms Matter in IPMAT
Across the last six IPMAT Indore papers (2020–2025), logarithm questions have appeared with the following frequency:
- Direct evaluation of log expressions — 1 to 2 questions every year
- Equation solving using log identities — 1 question almost every alternate year
- Combination with exponents/surds — embedded inside algebra or sequences problems
- Short-answer (TITA) form — 2024 and 2025 both featured a non-MCQ log question
That works out to 2–4 marks every paper from a topic where the entire theory fits on one A4 sheet. For an exam where the cutoff hovers between 60 and 70 percent, choosing not to prepare logarithms is choosing to give up a high-yield, low-effort area.
Definition and First Principles
If ax = N where a > 0 and a ≠ 1, then we write loga(N) = x. The number a is called the base, N is the argument, and x is the value of the logarithm.
Two non-negotiable conditions for a logarithm to exist:
- The argument
Nmust be strictly positive (N > 0) - The base
amust be positive and not equal to 1 (a > 0, a ≠ 1)
If the IPMAT question gives you loga(2 - x) and asks for the domain, the answer comes purely from 2 - x > 0, i.e., x < 2. Domain questions are gift marks once you internalise the existence conditions.
Common-base shorthand
When the base is 10 (called the common log), it is usually written without the base, i.e., log(N) means log10(N). When the base is e ≈ 2.718 (the natural log), it is written as ln(N). IPMAT typically sticks to base 10 and base 2, with one or two ln-based questions per cycle.
The Five Identities You Must Memorise
These are the only five rules you need. Every IPMAT log question is a re-skin of these.
| # | Identity | Statement | Quick Example |
|---|---|---|---|
| 1 | Product | loga(MN) = loga(M) + loga(N) | log(6) = log(2) + log(3) |
| 2 | Quotient | loga(M/N) = loga(M) − loga(N) | log(5) = log(10) − log(2) = 1 − 0.301 |
| 3 | Power | loga(Mk) = k · loga(M) | log(8) = 3·log(2) |
| 4 | Change of Base | loga(N) = logb(N) / logb(a) | log2(7) = log(7)/log(2) |
| 5 | Reciprocal | loga(b) · logb(a) = 1 | log3(5) · log5(3) = 1 |
Three corollaries you will use constantly
loga(a) = 1— log of base to itself is 1loga(1) = 0— log of 1 in any valid base is 0aloga(N) = N— exponentiating reverses the log
Standard Numerical Values to Memorise
IPMAT problems often demand decimal estimates without a calculator. Memorising eight values shortens your solution time dramatically:
| log10(N) | Decimal |
|---|---|
| log 2 | 0.301 |
| log 3 | 0.477 |
| log 5 | 0.699 (= 1 − log 2) |
| log 6 | 0.778 (= log 2 + log 3) |
| log 7 | 0.845 |
| log 8 | 0.903 (= 3·log 2) |
| log 9 | 0.954 (= 2·log 3) |
| ln 2 | 0.693 |
Six Question Archetypes IPMAT Loves
Archetype 1: Direct Evaluation
Compute log2(64) − log3(81). Answer: 6 − 4 = 2. These questions test whether you can express the argument as a power of the base.
Archetype 2: Equation Solving
Solve log2(x) + log2(x − 6) = 4. Use the product rule: log2[x(x−6)] = 4, so x² − 6x = 16, giving x = 8 or x = −2. Reject x = −2 (domain), so x = 8.
Archetype 3: Number of Digits
How many digits does 250 have? Compute 50·log 2 = 50·0.301 = 15.05. Take the integer part and add 1: 15 + 1 = 16 digits.
Archetype 4: Comparison without a Calculator
Which is larger, log3(7) or log5(11)? Use change of base: log3(7) = 0.845/0.477 ≈ 1.77 and log5(11) = 1.041/0.699 ≈ 1.49. So log3(7) wins.
Archetype 5: Nested or Chained Expressions
Evaluate log2(log3(81)). Inner: log3(81) = 4. Outer: log2(4) = 2.
Archetype 6: TITA-style Cleverness
If logx(2) = a and logx(3) = b, find logx(72). Decompose: 72 = 2³·3², so the answer is 3a + 2b. These show up in IPMAT 2024 and the IPM mock test paper of 2025 — a steady source of 4-mark questions.
Common Traps and How to Avoid Them
- Negative logs:
log(0.5)is negative, not undefined. Don’t reject negative answers in equations until you check the domain. - log(M+N) ≠ log M + log N — only product separates, never sums. This is the most common silly mistake in IPMAT mocks.
- Base of 1 or 0:
log1(N)is undefined; reject any equation that forces base = 1. - Implicit base: when no base is written, assume base 10 in IPMAT (not
e). Indian textbooks lean to common log. - Forgetting domain: when solving, always plug your answer back to confirm the argument is positive.
25-Question Practice Set with Solutions
Set a 30-minute timer. Aim for at least 20/25 to be IPMAT-ready.
Section A: Evaluation (Q1–Q8)
Q1. Find log5(125).
Solution: 125 = 5³, so answer is 3.
Q2. Compute log2(32) + log3(27).
Solution: 5 + 3 = 8.
Q3. Evaluate log4(64) − log9(81).
Solution: 3 − 2 = 1.
Q4. What is log10(0.001)?
Solution: 0.001 = 10−3, so answer is −3.
Q5. Find log√2(8).
Solution: 8 = 2³ = (√2)6, so answer is 6.
Q6. Evaluate log2(log2(16)).
Solution: Inner = 4, Outer = 2. Answer: 2.
Q7. Compute log3(54) − log3(2).
Solution: log3(27) = 3.
Q8. Find log5(0.04).
Solution: 0.04 = 1/25 = 5−2, answer is −2.
Section B: Equations (Q9–Q15)
Q9. Solve log2(x) = 5.
Solution: x = 25 = 32.
Q10. Solve logx(81) = 4.
Solution: x4 = 81 ⇒ x = 3.
Q11. Solve log3(x) + log3(x − 2) = 1.
Solution: x(x−2) = 3 ⇒ x² − 2x − 3 = 0 ⇒ x = 3 or −1. Domain rejects −1, so x = 3.
Q12. If log10(x²) = 4, find x.
Solution: x² = 10000 ⇒ x = ±100.
Q13. Solve 2·log5(x) − log5(2x − 1) = 0.
Solution: log5(x²/(2x−1)) = 0 ⇒ x² = 2x − 1 ⇒ (x−1)² = 0 ⇒ x = 1.
Q14. Solve log2(x − 1) + log2(x + 1) = 3.
Solution: log2(x² − 1) = 3 ⇒ x² − 1 = 8 ⇒ x = 3 or −3. Domain rejects −3, so x = 3.
Q15. Solve logx(2) · log2x(2) = log4x(2). (Hint: substitute y = log2(x).)
Solution: Cross-multiply via change of base. After simplification, y² + 3y − 2 = 0, giving x = 2(−3+√17)/2. (TITA-friendly.)
Section C: Number of Digits & Comparison (Q16–Q20)
Q16. How many digits in 2100?
Solution: 100·0.301 = 30.1 ⇒ 31 digits.
Q17. How many digits in 350?
Solution: 50·0.477 = 23.85 ⇒ 24 digits.
Q18. Which is greater: log2(5) or log3(10)?
Solution: log2(5) ≈ 2.32, log3(10) ≈ 2.10. So log2(5).
Q19. Find the characteristic of log(45000).
Solution: 45000 = 4.5 × 104, characteristic = 4.
Q20. If log 2 = 0.301, find log 50.
Solution: log 50 = log(100/2) = 2 − 0.301 = 1.699.
Section D: Mixed and Decomposition (Q21–Q25)
Q21. If log 2 = a and log 7 = b, find log 280.
Solution: 280 = 2³·5·7 = 2³·(10/2)·7, so answer = 3a + 1 − a + b = 2a + b + 1.
Q22. Simplify log3(5) · log5(7) · log7(9).
Solution: Telescope via change of base = log3(9) = 2.
Q23. If loga(b) = 4, find logb(a²).
Solution: logb(a²) = 2·logb(a) = 2/4 = 0.5.
Q24. Solve for x: xlog x = 100x.
Solution: Take log: (log x)² = 1 + log x. Let y = log x: y² − y − 1 = 0 ⇒ y = (1±√5)/2. Two solutions: x = 10(1+√5)/2 or 10(1−√5)/2.
Q25. If 3log3(7) = ?.
Solution: By the inverse identity, answer = 7.
Two-Week Drill Plan
| Day | Focus | Outcome |
|---|---|---|
| 1–2 | Memorise 5 identities + 8 decimal values | Formula sheet recall in under 90 seconds |
| 3–4 | Q1–Q8 evaluation drills (this set) | 2 minutes per question |
| 5–6 | Q9–Q15 equations | Domain checking automatic |
| 7–8 | Q16–Q20 digits + comparisons | Recall log 2, log 3, log 7 instantly |
| 9–10 | Q21–Q25 mixed decomposition | Express any number in base 2/3/5/7 |
| 11–12 | 30 fresh problems from prior IPMAT papers | Sub-2-minute solving pace |
| 13 | Diagnostic mock — 10 log questions in 12 min | ≥80% accuracy |
| 14 | Error log review + repeat weak archetypes | Move to algebra/quadratics next |
Frequently Asked Questions
Q. Are logarithm questions in IPMAT calculator-allowed?
No. IPMAT Indore, JIPMAT and IPMAT Rohtak are all non-calculator exams. Memorising the 8 decimal values listed above is essential for digit-counting and comparison questions.
Q. How many log questions appear in IPMAT Indore?
On average, 1–2 questions per quantitative paper. In 2024 and 2025, both papers had at least one log-based TITA question worth 4 marks.
Q. Do I need to know natural logarithms (ln) for IPMAT?
Helpful but not critical. The vast majority of questions use base 10 or base 2. One ln question every 2–3 cycles is the realistic frequency.
Q. What is the toughest log archetype in IPMAT?
Substitution-based equations like Q15 and Q24 in this set, where you reduce a log equation to a quadratic. Drill 10 of these and you will recognise the substitution pattern instantly.
Q. Where does this topic connect to other IPMAT chapters?
Logs feed directly into compound interest, population growth (sequences and series), and even data interpretation when the answer choices involve large powers. It is genuinely high-leverage prep.
Continue Your IPMAT Quant Drill
- IPMAT Number System 2027 — Divisibility, LCM/HCF, Remainders
- IPMAT Algebra 2027 — Linear, Quadratic, Progressions
- IPMAT Geometry 2027 — Triangles, Circles, Mensuration
- IPMAT Quant Strategy 2027 — 30-Day Plan
- IPMAT 2027 Preparation Timeline — Month-wise Plan
Diagnostic Quiz — 10 Questions, 12 Minutes
Take this test under timed conditions. Above 7/10 means logarithms are exam-ready; below 6/10 means re-drill Section B and Section D from the 25-question set above.
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