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NEET / JEE · Complete Reference

All Logarithm Rules for NEET

Every logarithm rule a NEET / JEE aspirant needs in one page — basic identities, the three operation rules, power and root rules, change-of-base, ln ↔ log conversion (the magic ×2.303), domain & sign rules, monotonicity, and the full chemistry / physics application set: pH, Arrhenius, Nernst, radioactive decay, ΔG°, buffer Henderson, Beer-Lambert and more. Read once cold, cover with your hand and recite. The three rules you'll reach for in 90 % of NEET problems are flagged at the bottom.

📏 1 · The Basics — what a logarithm is

A logarithm is just a tidy way to write the question "what power do I raise this number to?". It is the inverse of an exponential.

If ax = N ⇔ loga N = x
🎈 Plain English

"What power of a gives N?" — that answer is the logarithm of N to base a. So log10 1000 = 3 means "10 multiplied by itself 3 times gives 1000".

1.1 Three parts of a log expression

1.2 The two bases NEET cares about

NameSymbolBaseWhere it appears
Common loglog N or log10 N10pH, pOH, pKa, Nernst, decibels
Natural logln N or loge Ne ≈ 2.718Arrhenius, radioactive decay, exponential growth

📏 2 · Basic Identities

RuleFormulaPlain meaning
Log of 1loga 1 = 0Any base to the power 0 is 1.
Log of the baseloga a = 1Base raised to power 1 is itself.
Log of a base-powerloga(ax) = xRead the exponent off directly — base "cancels".
Inverse identityaloga N = NPower and log are inverses — they undo each other.
One-to-oneIf loga x = loga y, then x = yDifferent inputs give different log values.
Equal bases & argumentsIf loga x = logb x & x ≠ 1, then a = bEqual logs at the same non-trivial argument force equal bases.
10log x = x and eln x = x. These two "cancellations" appear in 1 out of every 4 NEET kinetics or pH questions.

📏 3 · The Three Operation Rules (most useful!)

RuleFormulaPlain meaning
Productlog(xy) = log x + log yMultiply two numbers → ADD their logs.
Quotientlog(x/y) = log x − log yDivide → SUBTRACT logs.
Reciprocallog(1/x) = − log xReciprocal of x flips the sign of log x.
Negative powerlog(x−n) = −n · log xNegative exponent flips the sign and pulls the coefficient out.
General product chainlog(x₁ x₂ x₃ …) = Σ log xᵢMultiplying any number of factors → sum the logs.
🎈 Plain English

Multiplication is hard, addition is easy. Logs turn multiplications INTO additions and divisions into subtractions. That's why slide-rules used to be the world's calculators before computers existed.

📏 4 · Power & Root Rules

RuleFormulaPlain meaning
Powerlog(xn) = n · log xPull the power out as a multiplier — the single most-used log rule in NEET kinetics.
Rootlog(n√x) = (1/n) · log xSame as power rule with fractional exponent 1/n.
Square rootlog √x = (1/2) log xSpecial case of the root rule.
Power in baselogan x = (1/n) · loga xRaise the BASE to a power → divide the log by n.
Power on both sideslogan xm = (m/n) · loga xCombine the two above.
📌 Example

log(8³) = 3 · log 8 = 3 · 3 log 2 = 9 × 0.30 = 2.71

📌 Example (power in base)

log9 27 = log(3³) = (3/2) · log3 3 = 3/2

4.1 Nested / Tower Powers — log(xyz) and log((xy)z)

These two look similar but they are different. The standard convention for exponentiation is right-associative, so xyz means x(yz), not (xy)z.

ExpressionWhat it meansApply log → answer
log(xyz)log(x(yz))  (tower)yz · log x
log((xy)z)log(xyz)  (rule of indices)yz · log x
log(xy · xz)log(xy+z)(y + z) · log x
log(xy/xz)log(xy−z)(y − z) · log x
🎈 The rule in one line

The power rule log(xn) = n · log x works no matter what shape n has. n can be a number (3), a variable (y), a product (yz), a sum (y + z), a power (yz), or even another logarithm. Whatever sits in the exponent comes out as the multiplier — full stop.

4.2 Step-by-step worked examples

📌 Example 1 — pure tower

log(232). The exponent is 3² = 9. So log(2⁹) = 9 · log 2 = 9 × 0.30 = 2.70.

Compare with log((2³)²) = log(2⁶) = 6 · log 2 = 6 × 0.30 = 1.80. Notice the two are completely different.

📌 Example 2 — symbolic tower

Simplify log(xyz).

Apply the power rule with n = yz: → yz · log x.

If you also want to crack yz, take a second log: log(yz · log x) = z · log y + log(log x). (You can keep peeling layers off.)

📌 Example 3 — variable in BOTH base & tower

Simplify log(xxx).

The exponent is xx. → xx · log x. (Then if you take log of the result you can split xx further as x · log x + log(log x).)

📌 Example 4 — bracketed power (different!)

Simplify log((x2)3).

Brackets force the rule of indices first: (x²)³ = x2·3 = x⁶.

log(x⁶) = 6 · log x.

📌 Example 5 — JEE-style logarithmic differentiation setup

Differentiate y = x(sin x).

Take log: log y = sin x · log x  (the power sin x comes out front).

Differentiate both sides: (1/y) · dy/dx = cos x · log x + sin x · (1/x).

So dy/dx = x(sin x)·[cos x · log x + sin x/x].

Common NEET / JEE trap: students write log(xyz) = yz · log x. That's wrong. yz is what you get from (xy)z only. The pure tower without brackets gives yz·log x.

4.3 What about logxy(xz)?

Use the "power in BOTH" rule (4.2 of section 4 main table):

logxy(xz) = (z/y) · logx x = z/y
📌 Example

log4 32 = log 2⁵ = 5/2 = 2.5.

📏 5 · Change of Base & Base Swapping

RuleFormulaPlain meaning
Change of baselogb x = loga x / loga bConvert from any base to any other — divide by the log of the new base.
Swap base ↔ argumentloga b = 1 / logb aSwap the base and the argument → take the reciprocal.
Product of swapsloga b · logb a = 1Direct consequence of swap rule.
Three-base chainloga b · logb c = loga c"Chain rule" — the middle base cancels.
n-base chainloga₁ a₂ · loga₂ a₃ · … · logaₙ₋₁ aₙ = loga₁ aₙGeneralisation of three-base chain.
📌 Example (change of base)

log2 32 = log 32 / log 2 = (5 log 2) / log 2 = 5  (since 25 = 32)

📏 6 · Natural Log ↔ Common Log Conversion

RuleFormulaPlain meaning
ln to logln x = 2.303 · log10 xThe single most-used conversion in chemistry kinetics — multiply log by 2.303.
log to lnlog10 x = 0.4343 · ln x= ln x / 2.303 = ln x · log e.
ln of exln(ex) = xNatural log and exponential cancel.
e to the lneln x = xThe other direction also cancels.
10 to the log10log x = x"Antilog of log of x is x."
ln 1 / ln e / ln 10ln 1 = 0, ln e = 1, ln 10 = 2.303Anchors.
2.303 is literally ln 10. Memorise it once — it appears in Arrhenius, Nernst, radioactive decay, ΔG° = −RT ln K conversion to −2.303 RT log K, and the integrated first-order rate law.

📏 7 · Sign & Domain Rules

SituationValue of log10 xWhy
x > 1Positivex is bigger than 10⁰ = 1.
x = 10Any base to power 0 = 1.
0 < x < 1NegativeNeed a negative power of 10 to shrink past 1.
x = 0Undefined (→ −∞)No finite power makes a positive base equal 0.
x < 0Undefined (in real numbers)No real exponent of a positive base gives a negative.
Base a ≤ 0 or a = 1Not allowedBase must be positive and ≠ 1.
🎈 Plain English

Logs only know how to handle positive arguments. Zero is the cliff edge (log → −∞), negatives fall off the cliff entirely (no real value).

📏 8 · Inequality (Monotonicity) Rules

If…Then…
Base a > 1 & x > y > 0loga x > loga y (log is increasing)
Base 0 < a < 1 & x > y > 0loga x < loga y (log is decreasing — flips!)
For any base, x = ylog x = log y
x = y means base & argument identicallog = 1
🎈 Plain English

For "normal" bases (10 or e), the log function is increasing: bigger input → bigger log. NEET / JEE always uses base > 1, so the second case rarely shows up.

📏 9 · Useful Derived Results

RuleFormula
Sum of log → log of productlog a + log b + log c = log(abc)
Coefficient of log → log of powern · log a = log(an)
Log of a quotient powerlog((a/b)n) = n · (log a − log b)
Log of a square root quotientlog √(a/b) = ½(log a − log b)
Log of factorial (large n)log 1 + log 2 + log 3 + … + log n = log(n!)
Log of m · √nlog(m·√n) = log m + ½ log n
Log of ekxln(ekx) = kx
Log of a · 10nlog(a × 10n) = log a + n (scientific notation)
⚡ The scientific-notation rule is gold for pH calculations: log(3.16 × 10⁻⁵) = log 3.16 − 5 ≈ 0.5 − 5 = −4.5 → pH = 4.5.

10 · Log Values You Must Memorise

Memorise three numbers (log 2, log 3, log 7) and you can compute almost every log NEET / JEE will ask.

Numberlog₁₀ valueDerivation
1010⁰ = 1
20.3010memorise
30.4771memorise
40.6020= 2 × log 2
50.6990= 1 − log 2 (since 5 = 10/2)
60.7781= log 2 + log 3
70.8451memorise
80.9030= 3 × log 2
90.9542= 2 × log 3
10110¹ = 10

10.1 Natural log values

ConstantValue
ln 20.693
ln 31.099
ln 51.609
ln 102.303
ln e1
e2.718

10.2 Antilog mental estimates

If log x ≈then x ≈
01
0.302
0.483
0.604
0.705
0.786
0.908
0.959
110
2100
31000
Characteristic and mantissa. Any log has two parts. log 250 = 2.398: the 2 (characteristic) comes from "digits before decimal − 1" — 250 has 3 digits → 2. The 0.398 (mantissa) is log 2.5.

🧪 11 · Chemistry Applications of Logarithms

Every NEET Chemistry chapter that involves "very large" or "very small" numbers uses logs.

11.1 Ionic Equilibrium

pH = − log[H⁺]    pOH = − log[OH⁻]    pKa = − log Ka    pKb = − log Kb
pH + pOH = 14   (at 25 °C)      pKa + pKb = 14
Henderson (acidic buffer): pH = pKa + log ([salt]/[acid])
Weak acid: pH = ½ (pKa − log C)

11.2 Chemical Kinetics

First-order: k = (2.303/t) · log([A]₀/[A])
Half-life (first-order): t½ = 0.693/k = ln 2 / k
Arrhenius: log(k₂/k₁) = (Ea/2.303 R) · (1/T₁ − 1/T₂)   |   ln(k₂/k₁) = (Ea/R) · (1/T₁ − 1/T₂)

11.3 Electrochemistry — Nernst

E = E° − (0.0591/n) · log Q  (at 298 K)
E° = (0.0591/n) · log K (at equilibrium)
For H₂ electrode: E = − 0.0591 · pH

11.4 Thermodynamics

ΔG° = − RT ln K = − 2.303 RT log K
At 298 K: ΔG° (kJ/mol) ≈ − 5.7 · log K

11.5 Beer-Lambert Spectroscopy

A = log(I₀/I) = ε · b · c
🧪 NEET trap

Absorbance is base-10 log, not natural log. Transmittance T = I/I₀ ⇒ A = − log T.

⚛️ 12 · Physics Applications of Logarithms

12.1 Modern Physics — Radioactive decay

N(t) = N₀ · e−λt   ⇒   ln(N/N₀) = − λt
t = (2.303/λ) · log(N₀/N)    t½ = ln 2 / λ = 0.693/λ

12.2 Sound — Intensity Level

β (dB) = 10 · log(I/I₀)
🎈 Plain English

The decibel scale is logarithmic because the ear handles intensities over a billion-fold range. +10 dB = 10× intensity; +20 dB = 100× intensity.

12.3 Optics & Spectroscopy

Absorbance A = log(I₀/I)
Magnitude scale (astronomy): m₁ − m₂ = − 2.5 log(F₁/F₂)

12.4 Misc physics derivations using logs

📝 13 · Worked NEET / JEE Examples

📌 Example 1 · pH from [H⁺]

If [H⁺] = 3.16 × 10⁻⁵, then pH = − log(3.16 × 10⁻⁵) = 5 − log 3.16 ≈ 5 − 0.5 = 4.5.

📌 Example 2 · Weak acid pH

0.1 M acetic acid, Ka = 1.8 × 10⁻⁵. pKa = − log(1.8 × 10⁻⁵) = 5 − log 1.8 ≈ 5 − 0.255 = 4.745. pH = ½(pKa − log C) = ½(4.745 + 1) = 2.87.

📌 Example 3 · Arrhenius numerical

k doubles from 300 K to 310 K. Find Ea.
log 2 = (Ea/2.303 R)(1/300 − 1/310). log 2 = 0.301. RHS factor = (10/300×310) ≈ 1.075×10⁻⁴. Ea = (0.301 × 2.303 × 8.314)/(1.075×10⁻⁴) ≈ 53.6 kJ/mol.

📌 Example 4 · Nernst

Cell with E° = 0.59 V and n = 1. log K = E°·n/0.0591 = 0.59/0.0591 = 10 → K = 10¹⁰.

📌 Example 5 · First-order half-life

k = 5 × 10⁻⁴ s⁻¹. t½ = 0.693/k = 0.693/(5 × 10⁻⁴) = 1386 s.

📌 Example 6 · ΔG° from K

K = 10⁵ at 298 K. ΔG° = − 2.303 RT log K = − 2.303 × 8.314 × 298 × 5 = − 28.5 kJ/mol. Shortcut: ΔG° (kJ) ≈ − 5.7 × log K = − 5.7 × 5 = −28.5 kJ/mol.

📌 Example 7 · Radioactive decay

Half-life 10 years. After 40 years, fraction remaining = (1/2)⁴ = 1/16. Equivalent log form: ln(N/N₀) = − λt = − 4 ln 2 → N/N₀ = e−4 ln 2 = 2−4 = 1/16. 1/16 of N₀.

📌 Example 8 · Buffer Henderson

Acetic acid + sodium acetate, both 0.1 M. pH = pKa + log(salt/acid) = pKa + log 1 = pKa = 4.74.

📌 Example 9 · Absorbance

If transmittance T = 0.1, then A = − log 0.1 = 1. If T = 0.01, A = 2. Each unit of A = 10× drop in transmitted light.

📌 Example 10 · Decibel

Doubling sound intensity adds 10 log 2 ≈ 3 dB. Ten-fold increase adds 10 log 10 = 10 dB. Hundred-fold adds 20 dB.

🔑 14 · One-Page Cheat Sheet

The Three Rules NEET Tests in 90% of Questions

RankRuleMost common use
1Product / Power   log(xy) = log x + log y, log(xn) = n log xpH, K calculations, kinetics
2ln ↔ log   ln x = 2.303 log xArrhenius, Nernst, radioactive decay
3Inverse identity   10log x = x, eln x = xantilog, finding [H⁺] from pH, finding N from ln(N/N₀)

Anchor values

ConstantValue
log 2 / log 3 / log 70.30 / 0.48 / 0.85
ln 2 / ln 100.693 / 2.303
0.0591 V= (2.303 RT/F) at 298 K — Nernst factor
0.693/kFirst-order half-life formula
5.7 kJ/mol= 2.303 RT at 298 K → ΔG° (kJ) ≈ −5.7 log K

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⚡ Read this card daily for a week before the exam. Page-1 Chemistry / Physics questions reward this single page more than 100 pages of mock tests.