Hello students, I am Rahul Sir from OdTutor, and over the years of training aspirants for IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, SSC, and Railway examinations, I have seen one topic consistently intimidate students who are otherwise quite well-prepared — Logarithm. The moment students see the word “log” in a question, something switches off in their brain. They skip it, mark it for later, and often never return to it during the exam, losing easy marks in the process.
Here is the truth I tell every batch of students I teach: Logarithm is one of the most formula-friendly chapters in the entire quantitative aptitude syllabus. It does not require lengthy calculations, it does not require complex reasoning, and once you understand what a logarithm actually means, the formulas feel logical rather than arbitrary. You stop memorizing and start understanding, and that makes all the difference.
At OdTutor, I have developed a step-by-step teaching approach for Logarithm that takes even the most hesitant student from absolute confusion to confident problem-solving within a matter of days. In this article, I am going to walk you through every important concept, formula, and question type that IBPS PO and Clerk exams test, with clear solved examples at every stage. Read it from beginning to end, practice the examples alongside, and I promise you — Logarithm will become one of your favorite scoring topics in the exam.
Let’s get started.
1. What Exactly Is a Logarithm? The Concept Explained Simply
Before any formula, any shortcut, or any trick, you must understand what the word “logarithm” actually means. I spend a full 20 minutes on this in my live classes, because every single formula that comes afterward is rooted in this one definition.
Consider this simple question: “To what power must we raise 2 to get 8?” You already know the answer — 2 raised to the power 3 gives 8, so the answer is 3. Logarithm is simply a way of writing this question and its answer in a formal mathematical notation.
We write it as: log₂(8) = 3
This is read as “log base 2 of 8 equals 3,” and it means exactly the same thing as 2³ = 8.
The general definition is: logₐ(b) = c means aᶜ = b
Here, ‘a’ is called the base, ‘b’ is the number whose logarithm is being taken, and ‘c’ is the answer — the exponent to which ‘a’ must be raised to get ‘b’.
I give my students this mental anchor: the logarithm answers the question “what is the exponent?” That’s all it is. Every time you see logₐ(b), your brain should immediately ask: “a raised to WHAT power gives b?” The answer to that question is your logarithm value.
This conversion between logarithmic form and exponential form is the very first skill you must master, because every property and formula of logarithm is simply the exponential rule rewritten in log language. Get this conversion absolutely smooth, and the rest of the chapter flows naturally.
2. Essential Properties of Logarithm You Must Know by Heart
Now that the definition is clear, let’s build the formula toolkit. These properties appear directly or indirectly in almost every Logarithm question in IBPS exams. I want you to not just memorize them, but understand why each one makes sense based on what you already know about exponents.
Property 1 — Product Rule: log(m × n) = log m + log n
When you multiply two numbers, their logs add. This mirrors the exponent rule: aˣ × aʸ = aˣ⁺ʸ.
Property 2 — Quotient Rule: log(m / n) = log m − log n
When you divide two numbers, their logs subtract. This mirrors: aˣ / aʸ = aˣ⁻ʸ.
Property 3 — Power Rule: log(mⁿ) = n × log m
An exponent inside a log comes out as a multiplier in front. This is extremely frequently tested.
Property 4 — Change of Base Rule: logₐ(b) = log(b) / log(a)
This lets you convert any log to base 10, which makes calculations far easier.
Property 5 — Log of 1: logₐ(1) = 0 for any base a
Because a⁰ = 1 always, the log of 1 is always 0 regardless of the base.
Property 6 — Log of Base Itself: logₐ(a) = 1
Because a¹ = a always, the log of the base equals 1.
Property 7 — Log Base Reciprocal: logₐ(b) = 1 / logᵦ(a)
Switching the base and the number gives you the reciprocal. This is a lifesaver in competitive exams.
Write all seven properties on a single sheet. Review them every morning for one week until you can write all seven from memory in under two minutes.
3. Solved Examples: Applying Basic Properties
Let’s immediately put those properties to work with examples that mirror the style of actual IBPS questions.
Example 1: Find the value of log₂(32).
Solution:
We need to find: 2 raised to what power gives 32?
Since 32 = 2⁵, we have log₂(32) = 5. That’s it.
Example 2: Simplify: log(6) + log(5) − log(3)
Solution:
Using Product Rule first: log(6) + log(5) = log(6 × 5) = log(30)
Then using Quotient Rule: log(30) − log(3) = log(30/3) = log(10) = 1
(Since log base 10 of 10 = 1, and when no base is written, base 10 is assumed.)
The answer is 1.
Example 3: Simplify: log₃(81) − log₃(9)
Solution:
log₃(81) = log₃(3⁴) = 4
log₃(9) = log₃(3²) = 2
Answer = 4 − 2 = 2
Alternatively using Quotient Rule: log₃(81/9) = log₃(9) = 2. Same answer.
Example 4: If log(2) = 0.3010, find log(8).
Solution:
log(8) = log(2³) = 3 × log(2) = 3 × 0.3010 = 0.9030
This is a hugely important question type. IBPS almost always gives you a standard log value and asks you to calculate another one by expressing it as a power of a given number. I will cover this pattern extensively in the coming sections.
4. The Standard Log Values You Must Memorize
In IBPS PO and Clerk exams, Logarithm questions almost always revolve around a few standard log values that you are either given in the question or expected to know. I cannot tell you how many students lose marks simply because they hadn’t memorized these before entering the exam hall.
Here are the values you absolutely must know:
log(1) = 0 (log of 1 is always 0)
log(2) ≈ 0.3010
log(3) ≈ 0.4771
log(5) ≈ 0.6990 (tip: since 2 × 5 = 10, log 5 = log 10 − log 2 = 1 − 0.3010 = 0.6990)
log(7) ≈ 0.8451
log(10) = 1
log(11) ≈ 1.0414
Now here’s the golden trick I teach at OdTutor: once you know log(2), log(3), and log(7), you can derive almost any other standard log value.
For example:
log(4) = log(2²) = 2 × 0.3010 = 0.6020
log(6) = log(2 × 3) = 0.3010 + 0.4771 = 0.7781
log(9) = log(3²) = 2 × 0.4771 = 0.9542
log(12) = log(4 × 3) = 0.6020 + 0.4771 = 1.0791
log(15) = log(3 × 5) = 0.4771 + 0.6990 = 1.1761
This derivation skill turns the memorization burden into a simple calculation exercise. You only need to truly memorize three or four base values, and everything else can be derived in under 10 seconds during the exam.
5. Finding Unknown Values Using Log Properties — A Key Question Pattern
One of the most commonly asked question patterns in IBPS exams involves finding the value of an expression when certain standard log values are given. Let’s solve several of these together.
Question: If log(2) = 0.3010 and log(3) = 0.4771, find the value of log(36).
Solution:
log(36) = log(4 × 9) = log(4) + log(9)
= log(2²) + log(3²)
= 2 × log(2) + 2 × log(3)
= 2 × 0.3010 + 2 × 0.4771
= 0.6020 + 0.9542
= 1.5562
Question: If log(2) = 0.3010, find the number of digits in 2²⁰.
Solution:
This is a slightly advanced but very important question type for IBPS PO.
log(2²⁰) = 20 × log(2) = 20 × 0.3010 = 6.020
The number of digits in any number N is given by: floor(log N) + 1
Here, floor(6.020) = 6
So number of digits = 6 + 1 = 7 digits
I always emphasize this “number of digits” formula separately in my classes because it’s a beautiful application of logarithm that IBPS PO prelims and mains both test regularly. The rule is simple: find the log of the number, take the integer part (characteristic), add 1, and that gives you the number of digits. Practice this until it feels automatic.
6. The Characteristic and Mantissa — Understanding Log Values Deeply
This section deals with concepts that are slightly more advanced but appear regularly in IBPS PO Mains and SBI PO exams. Understanding characteristic and mantissa will also help you solve “number of digits” problems more confidently.
Every logarithm value (in base 10) has two parts:
Characteristic: The integer part (the part before the decimal point).
Mantissa: The decimal part (the part after the decimal point), always non-negative.
For example, log(250) = log(2.5 × 100) = log(2.5) + log(100) = 0.3979 + 2 = 2.3979
Here, characteristic = 2 and mantissa = 0.3979
Key Rules for Characteristic:
For a number greater than or equal to 1: the characteristic = (number of digits − 1)
So for 250, which has 3 digits, characteristic = 3 − 1 = 2. ✓
For a number between 0 and 1 (a decimal): the characteristic is negative and equals −(number of zeros after decimal point before the first significant digit + 1). This rule applies to questions involving small decimal numbers.
Why does this matter in exams?
Because many IBPS questions ask: “Find the characteristic of log(N)” or “The logarithm of a number has characteristic 3. What can you say about the number of digits in that number?”
If characteristic = 3, then number of digits = 3 + 1 = 4, meaning the number is between 1000 and 9999.
I always tell students: characteristic tells you the order of magnitude of a number. Mantissa tells you the “fingerprint” of the significant digits. Together, they give you the complete picture of the number in logarithmic language. Spend time understanding this and you’ll find “number of digits” problems become completely trivial to solve.
7. Change of Base Formula and Its Applications
The Change of Base formula is one of the most powerful tools in the logarithm toolkit, especially for IBPS PO exam questions where the base is unusual or the question involves multiple different bases.
Formula: logₐ(b) = log(b) / log(a) = ln(b) / ln(a)
This formula allows you to convert any logarithm into base 10, which is much easier to handle.
Example 1: Find the value of log₈(64).
Method 1 (Direct): 8² = 64, so log₈(64) = 2
Method 2 (Change of Base): log(64) / log(8) = log(2⁶) / log(2³) = 6 log(2) / 3 log(2) = 6/3 = 2 ✓
Example 2: Evaluate: log₄(8) × log₉(27)
Solution:
log₄(8) = log(8)/log(4) = log(2³)/log(2²) = 3 log(2) / 2 log(2) = 3/2
log₉(27) = log(27)/log(9) = log(3³)/log(3²) = 3 log(3) / 2 log(3) = 3/2
Product = 3/2 × 3/2 = 9/4 = 2.25
A crucial trick from my class: Whenever you see a logarithm expression where both the number and the base are powers of the same smaller number, immediately convert using powers. For example, log₄(8): both 4 and 8 are powers of 2, so convert both — 4 = 2², 8 = 2³. Then log₄(8) = 3/2 almost instantly. This trick makes an otherwise 4-step process happen in one line and saves precious seconds in the exam.
8. Solving Logarithmic Equations
Some IBPS PO questions and many SBI PO questions present logarithm in equation form, where you must find the value of an unknown variable. Let me walk you through the standard types.
Type 1: Simple Log Equation
Question: If log(x) = 2, find x.
Solution: log(x) = 2 means 10² = x, so x = 100.
Type 2: Equation With Properties
Question: If log(x − 1) + log(x + 1) = log(8), find x.
Solution:
Using Product Rule: log[(x−1)(x+1)] = log(8)
Since the logs are equal and have the same base: (x−1)(x+1) = 8
x² − 1 = 8
x² = 9
x = 3 (we take positive value since log requires positive arguments)
Type 3: Chain Equation
Question: If log₂(log₃(x)) = 1, find x.
Solution:
log₂(log₃(x)) = 1 means 2¹ = log₃(x), so log₃(x) = 2
Now, log₃(x) = 2 means 3² = x, so x = 9.
For all equation types, I teach my students one golden approach: always convert the log equation into its exponential form as the very first step. The moment you write “log = something” as “base^something = number,” the equation becomes a familiar algebraic problem and the fear disappears. Students consistently tell me this one mindset shift alone — thinking in exponential form — transforms their comfort with logarithm equations entirely.
9. Tricky Shortcut Questions That Appear in IBPS Exams
Let me now walk you through some of the cleverly crafted questions that IBPS examiners love to include, and the shortcuts I teach my students to crack them in under 45 seconds.
Shortcut Type 1: logₐ(b) × logᵦ(c) = logₐ(c)
This is called the “chain rule” of logarithms, and it’s a huge time saver.
Question: Find the value of log₂(3) × log₃(4) × log₄(32).
Solution:
Using the chain rule step by step:
log₂(3) × log₃(4) = log₂(4) = log₂(2²) = 2
Now, 2 × log₄(32) = log₄(32) … wait, let me redo this properly.
log₂(3) × log₃(4) × log₄(32) = log₂(32) = log₂(2⁵) = 5
Shortcut Type 2: 1/logₙ(a) = logₐ(n)
Question: Find: 1/log₂(100) + 1/log₃(100) + 1/log₅(100)
Solution:
1/log₂(100) = log₁₀₀(2)
1/log₃(100) = log₁₀₀(3)
1/log₅(100) = log₁₀₀(5)
Sum = log₁₀₀(2 × 3 × 5) = log₁₀₀(30)
= log(30)/log(100) = log(30)/2
= (log 3 + log 10)/2 = (0.4771 + 1)/2 = 1.4771/2 = 0.7386
These shortcut patterns appear regularly, especially in IBPS PO Mains. Recognizing the pattern instantly — rather than working through each step from scratch — is what separates students who score 18 out of 20 in quant from those who score 12. Pattern recognition is a skill that comes from deliberate, structured practice, not just random solving.
10. Practice Strategy for Mastering Logarithm Before the Exam
Let me end this article the same way I close every classroom session — with a clear, honest, actionable plan for what to do next.
Step 1 — Foundation (Days 1–3): Spend your first three days purely on understanding the definition and the seven properties listed in section 2. Don’t touch a single practice problem yet. Read the definitions, understand where each property comes from, write them out from memory repeatedly, and connect each log property to its corresponding exponent rule. This foundation phase is non-negotiable.
Step 2 — Standard Values (Day 4): Memorize log(2), log(3), log(7), and derive all other standard values from these. Then practice deriving them from scratch without looking at notes until you can do any standard log value in under 10 seconds.
Step 3 — Basic Questions (Days 5–7): Solve only direct, formula-application questions — similar to those in sections 3, 4, and 5 of this article. The goal here is accuracy, not speed. Check every answer, understand every error.
Step 4 — Intermediate Questions (Days 8–12): Move to characteristic and mantissa problems, change of base questions, and simple log equations as covered in sections 6, 7, and 8.
Step 5 — Advanced Shortcuts (Days 13–15): Practice the chain rule, reciprocal shortcuts, and the tricky pattern questions from section 9. These require pattern recognition that only comes from seeing enough examples.
Step 6 — Mixed Timed Practice (Days 16–20): Set 45 seconds per question and solve mixed Logarithm question sets. Include Logarithm in your daily mock test practice so it appears alongside other quant topics, just as it would in the real exam.
Step 7 — Error Log Review (Ongoing): Every wrong answer deserves a written explanation in your error log — was it a wrong formula, a calculation slip, or a misread question? Reviewing this weekly is how you steadily eliminate your personal blind spots and build real exam-day confidence.
Logarithm rewards consistent, structured effort more than raw intelligence. I have seen average students outperform brilliant ones in this topic simply because they followed a disciplined practice schedule. Be that student.
How Teachers from OdTutor Can Help
At OdTutor, our experienced trainers like Rahul Sir break down intimidating topics like Logarithm into simple, digestible steps that any student can follow, regardless of their mathematics background. Through live concept classes, doubt-clearing sessions, curated practice sheets, and full-length mock tests mapped to actual IBPS PO and Clerk exam patterns, OdTutor ensures that you don’t just understand Logarithm once, but can reliably solve any question under exam pressure. With personalized feedback on your performance, small focused batches, and a teaching approach built around shortcuts and pattern recognition, OdTutor gives you the structured guidance needed to turn your weakest topic into your most dependable scoring area.