A logarithm answers one simple question: to what power must this base be raised to get this number?
When you write log_b(a) = x, it means b^x = a. That's the entire definition. Everything else — all the rules, all the tricks — is a consequence of this one equivalence.
Here's an analogy that sticks. Suppose you know that 10^3 = 1000. Now someone asks: "what power of 10 gives 1000?" The answer is 3. That answer — 3 — is log₁₀(1000). Logarithm is just the inverse operation to exponentiation, the same way subtraction is the inverse of addition.
Two bases you must be comfortable with:
log₁₀, written as log): used in most NDA problems. log 1 = 0, log 10 = 1, log 100 = 2.ln, base e ≈ 2.718): appears less in NDA Maths directly, but the properties are identical.Domain restrictions — this is where most marks are lost, not gained. log_b(a) is defined only when:
a > 0 (argument must be strictly positive)b > 0 and b ≠ 1 (base must be positive and not equal to 1)Violate either condition and the expression is undefined. In equation-solving questions, any solution that makes an argument non-positive or zero must be rejected. This is not a technicality — NDA has tested this directly.
Antilog is simply the reverse: antilog_b(x) = b^x. If log₁₀(x) = 2, then antilog₁₀(2) = 10^2 = 100 = x.
Think of logs and antilogs as a pair of doors — one takes you from the exponent world into the log world, the other brings you back.
These four identities are non-negotiable. Derive them once from the definition, then use them without hesitation.
1. Product Rule: log_b(mn) = log_b(m) + log_b(n)
Why it works: If b^p = m and b^q = n, then mn = b^p · b^q = b^(p+q). So log_b(mn) = p + q = log_b(m) + log_b(n).
2. Quotient Rule: log_b(m/n) = log_b(m) − log_b(n)
Same logic: m/n = b^p / b^q = b^(p−q).
3. Power Rule: log_b(m^k) = k · log_b(m)
If b^p = m, then m^k = b^(kp), so log_b(m^k) = kp = k · log_b(m).
4. Change of Base: log_b(a) = log_c(a) / log_c(b)
This is the most underused property in exam halls. It lets you convert any log to base 10 (or any convenient base) using a calculator — or in NDA's case, lets you manipulate expressions algebraically.
A critical corollary: log_b(a) · log_a(b) = 1, which means log_b(a) = 1 / log_a(b). This appears constantly in NDA problems — memorise this reciprocal relationship.
| Expression | Value | Why |
|---|---|---|
| log_b(1) | 0 | because b^0 = 1 always |
| log_b(b) | 1 | because b^1 = b always |
| log_b(b^k) | k | direct from power rule |
| b^(log_b(a)) | a | logs and exponentials cancel |
When you see a log equation, your first two moves are:
log_b(LHS) = log_b(RHS) with the same base, then cancel the logs to get LHS = RHS.For equations like log₄(x−1) = log₂(x−3), the bases differ. The move is change of base: log₄(x−1) = log₂(x−1) / log₂(4) = log₂(x−1) / 2. Now both sides share base 2 and you can proceed.
Here's where students drop marks silently. When you remove the log from an inequality:
b > 1: the inequality direction stays the same. log_b(a) > log_b(c) means a > c.0 < b < 1: the inequality flips. log_b(a) > log_b(c) means a < c.Think of it this way: log_{0.5} is a decreasing function, so larger log value means smaller argument.
t + 1/t PatternLook at the PYQ on log_{sin x}(cos x) + log_{cos x}(sin x) = 2. If you let t = log_{sin x}(cos x), then log_{cos x}(sin x) = 1/t (by the reciprocal property). The equation becomes t + 1/t = 2.
Now, t + 1/t = 2 has exactly one solution: t = 1. This is because t + 1/t − 2 = (t−1)²/t = 0 implies t = 1.
This t + 1/t pattern — where log and its reciprocal appear together — is a recurring NDA structure. Recognise it and you save 90 seconds.
t + 1/tFor t > 0: by AM-GM inequality, (t + 1/t)/2 ≥ √(t · 1/t) = 1, so t + 1/t ≥ 2.
This means t + 1/t can never be less than 2 for positive t. For t < 0: t + 1/t ≤ −2.
The PYQ on k = 2 − t − 1/t exploits exactly this. Since t + 1/t ≥ 2, we get k ≤ 0. So k = 1 is impossible.
When you see multiplication inside a log, it becomes addition outside. When you see division inside, it becomes subtraction outside. The trick: logs "hate" multiplication and division — they break them apart into addition and subtraction.
Micro-example: log(50) = log(100/2) = log 100 − log 2 = 2 − 0.301 = 1.699.
Standard method (recalling log 50 from memory): 15-20 seconds if you remember it. This breakdown: 5 seconds. Saves you 10-15 seconds per problem.
log_b(a) = 1 / log_a(b) — the base and argument swap, and the whole thing flips.
Use this whenever a problem has log_b(a) and log_a(b) appearing in the same expression. Let t = log_b(a), then log_a(b) = 1/t, and the expression becomes t + 1/t (or t − 1/t, or t · 1/t = 1).
Micro-example: log_{sin x}(cos x) + log_{cos x}(sin x) = 2 becomes t + 1/t = 2 in one step versus three steps of change-of-base manipulation. Saves 2 algebraic steps.
When an equation has terms like 5^x + 4^x + 3^x + 2^x + 1 inside a log and the options include small integers (especially 0 or 1), test x = 0 first — every exponential term becomes 1.
Micro-example: 5^0 + 4^0 + 3^0 + 2^0 + 1 = 1+1+1+1+1 = 5 and you need the argument to equal 5. Done.
Standard approach (algebraic manipulation): 60-90 seconds. Substitution: 10 seconds. Use it whenever you see mixed exponentials inside a log.
When two log terms have different bases where one base is a power of the other (like base 4 and base 2), convert to the common smaller base using log_{b^k}(x) = (1/k) log_b(x).
Micro-example: log₄(x−1) = (1/2)log₂(x−1). Now both sides have base 2. Apply the one-to-one property and solve algebraically.
This reduces a "different bases" problem to a "same base" problem in one step. Saves the mental overhead of full change-of-base formula.
For any t > 0, t + 1/t ≥ 2. Use this to immediately determine the range of expressions of the form a − t − 1/t or a + t + 1/t without solving.
Micro-example: k = 2 − t − 1/t. Since t + 1/t ≥ 2, immediately k ≤ 0. Any answer option with k > 0 is ruled out in under 5 seconds — no algebra needed.
Standard approach (trying specific values of t for each option): 40-60 seconds. AM-GM observation: 5 seconds.
When you see a logarithm question in the NDA exam, run through this decision tree:
Step 1 — Identify the structure.
log_b(a) and log_a(b) in the same expression? Use the reciprocal property immediately. Set t = log_b(a).Step 2 — Simplify both sides. Apply product, quotient, and power rules to reduce each side to a single log expression.
Step 3 — Solve the resulting equation.
Once log_b(LHS) = log_b(RHS), equate LHS = RHS.
Step 4 — Check domain before marking your answer. Substitute back into every log argument. Any argument ≤ 0 disqualifies that solution. This step eliminates traps set specifically for careless solvers.
Step 5 — If the expression asks for a range or "which value is impossible":
Use AM-GM or known inequalities on t + 1/t to bound the expression without computing specific values.
Why this question: Tests whether you can convert a log expression on the LHS into a recognizable log value before handling the complex RHS.
Solving path: Left side — 1 − log₁₀ 2 = log₁₀ 10 − log₁₀ 2 = log₁₀(10/2) = log₁₀ 5. So the argument of the log on the right must equal 5. Test x = 0: 5^0 + 4^0 + 3^0 + 2^0 + 1 = 5. Confirmed. No algebra beyond this is needed.
Why this question: Tests the reciprocal property of logs — a pattern that recurs across NDA sessions. Students who don't recognise log_{cos x}(sin x) = 1 / log_{sin x}(cos x) spend 2-3 minutes on algebra that takes 30 seconds with the right substitution.
Solving path: Let t = log_{sin x}(cos x). Then log_{cos x}(sin x) = 1/t. Equation becomes t + 1/t = 2, i.e., t² − 2t + 1 = 0, i.e., (t−1)² = 0, so t = 1. This means cos x = sin x, so tan x = 1. Smallest positive solution: x = π/4.
Why this question: Tests whether you can convert log statements into exponential form and then reconstruct the combined expression. The key move is recognising ace = (bd²f³)^p.
Solving path: From log_b a = p, write a = b^p. From log_d c = 2p, write c = d^{2p}. From log_f e = 3p, write e = f^{3p}. Multiply: ace = b^p \cdot d^{2p} \cdot f^{3p} = (bd^2f^3)^p. Raise both sides to 1/p: (ace)^{1/p} = bd^2f^3.
Why this question: Two logs with different but related bases, leading to a quadratic after change of base — and then a domain-check trap. Missing the domain step costs the question.
Solving path: log_4(x−1) = (1/2)log_2(x−1). Equation becomes (1/2)log_2(x−1) = log_2(x−3), so log_2(x−1) = 2\log_2(x−3) = log_2(x−3)^2. Equate arguments: x−1 = (x−3)^2 = x^2 − 6x + 9, giving x^2 − 7x + 10 = 0, so (x−2)(x−5) = 0. Solutions: x = 2 or x = 5. Check x = 2: argument of log_2(x−3) = log_2(−1) — undefined. Reject. Only x = 5 is valid. Answer: one solution.
Why this question: Tests the AM-GM bound on log expressions — a purely reasoning-based question where no equation needs to be solved, only bounded.
Solving path: Let log_x y = t. Since x ≥ y > 1, we have 0 < t ≤ 1. Then log_x(x/y) = 1 − t and log_y(y/x) = 1 − \log_y x = 1 − 1/t. So k = (1−t) + (1−1/t) = 2 − t − 1/t. For t \in (0, 1], by AM-GM: t + 1/t \geq 2, so k \leq 0. The value k = 1 > 0 is impossible.
Not checking the domain after solving. Quadratic solutions from log equations routinely include one invalid root. Always substitute back and verify every log argument is strictly positive before marking an answer.
Treating log(a + b) as log a + log b. The product rule says log(ab) = log a + log b. There is no simplification for log(a + b) in general. This confusion costs marks every session.
Forgetting that log_b(a) = 1/log_a(b) — not log_a(b). The reciprocal goes on the entire fraction, not just the argument. If log_2(8) = 3, then log_8(2) = 1/3, not the other way around.
Applying log to both sides of an equation without checking sign. You can only take log of both sides if both sides are strictly positive. If the problem involves variables, verify positivity before applying log.
Ignoring the base constraint in inequalities. For 0 < b < 1, the log function is decreasing, so the inequality flips when you remove the log. Many students miss this and get the inequality direction backwards.
Assuming log₄(x) = 2 log₂(x). The correct conversion is log_4(x) = (1/2)log_2(x), not 2 log_2(x). Check the conversion: log_{b^k}(x) = (1/k) \log_b(x).