Important Logarithms Formulas for JEE and NEET

What are Logarithms?

A logarithm expresses the power to which a number (the base) must be raised to obtain another number. It provides a compact way to represent very large or very small numbers and converts multiplicative relationships into additive ones, which is useful in many branches of science, engineering and competitive examinations.

The formal definition is:

If b is a positive real number not equal to 1 and a is a positive real number, b^y = a ⇔ log _b a = y.

• a and b are positive real numbers.
• b ≠ 1.
• y is a real number.
• a is called the argument (the quantity inside the log).
• b is called the base (the small subscript of the log).

In plain words: a logarithm answers the question "How many times is the base multiplied by itself to get the argument?"

Example: How many 3's must be multiplied to obtain 27?

3 × 3 × 3 = 27, so 3 is used three times. Hence:

log₃(27) = 3

Logarithm (basic relation)

Let a, b be positive real numbers and a ≠ 1. If a^x = b then

log_ab = x

Example:

3⁵ = 243 ⇔ log₃243 = 5

Types of Logarithms

• Natural logarithm: base e (≈ 2.71828...). Written as ln N; that is ln N = log_eN. Example: ln 7 = log_e7.
• Common (Brigg's) logarithm: base 10. Written as log₁₀N. Example: log₁₀100 = 2.

Logarithm Properties (Fundamental)

• log_a1 = 0, for a > 0 and a ≠ 1.
• log_aa = 1, for a > 0 and a ≠ 1.
• log_aa^x = x, for real x.
• a^log_ax = x, for x > 0.
• Product rule: log_a(mn) = log_am + log_an, for m,n > 0.
• Quotient rule: log_a(m/n) = log_am - log_an, for m,n > 0.
• Power rule: log_a(mⁿ) = n·log_am, for m > 0.
• log_a(1/m) = - log_am, for m > 0.
• 
• If log_ab = x (a, b > 0, a ≠ 1), then:
  • log_1/ab = -x
  • log_a(1/b) = -x
  • log_1/a(1/b) = x
• Change of base relation: log_a^mb = (1/m)·log_ab.
• Monotonicity: log_ax is decreasing if 0 < a < 1, and increasing if a > 1.
• If 0 < a < 1 and b,c > 0 then:
  • log_ab ≥ log_ac ⇔ b ≤ c
  • log_ab ≥ c ⇔ b ≤ a^c
• If a > 1 and b,c > 0 then:
  • log_ab ≥ log_ac ⇔ b ≥ c
  • log_ab ≥ c ⇔ b ≥ a^c

Characteristics and Mantissa

Characteristic: the integral (integer) part of a logarithm.

Mantissa: the fractional (decimal) part of a logarithm; always non-negative in common use.

Example: log 3274 = 3.5150 → characteristic = 3, mantissa = 0.5150.

Points to remember about characteristic (common logarithm)

• The characteristic of the common logarithm of a positive number less than 1 is negative.
• The characteristic of the common logarithm of a positive number greater than 1 is positive (or zero when the argument is 1).
• If log₁₀x = n.abcd (characteristic n), then the number of integral values x can take between consecutive powers of 10 is 10ⁿ⁺¹ - 10ⁿ.
• If the characteristic of log₁₀x is negative (-n), then the number of zeros between the decimal point and the first non-zero digit of x is (n - 1).

Important Conversions (Inequalities & comparisons)

• For a > 1, a^b > c (with c > 0) ⇔ log_ac < b.
• For 0 < a < 1, a^b > c (with c > 0) ⇔ log_ac > b.

Logarithm Rules (Summary and calculus relations)

The principal algebraic rules are the Product, Quotient and Power rules, together with the change-of-base and base-switch relations. For calculus, derivatives and integrals of logarithms are often used.

• Product rule: log_b(mn) = log_bm + log_bn
• Division rule: log_b(m/n) = log_bm - log_bn
• Power (exponential) rule: log_b(mⁿ) = n·log_bm
• Change of base: log_bm = log_am / log_ab
• Base switch: log_ba = 1 / log_ab
• Derivative: if f(x) = log_b(x) then f′(x) = 1 / (x·ln b)
• Integral: ∫ log_b(x) dx = x·(log_b(x) - 1/ln b) + C

Useful compact identities (always helpful in manipulations):

• log_b(mn) = log_bm + log_bn
• log_b(m/n) = log_bm - log_bn
• log_b(x^y) = y·log_bx
• m·log_bx + n·log_by = log_b(x^myⁿ)
• log_b(m + n) = log_bm + log_b(1 + n/m) (useful when n << m)
• log_b(m - n) = log_bm + log_b(1 - n/m) (valid when m > n)

Solved Examples

Question 1: If log₂X + log₄X = log_0.25√6 and x > 0, then x is:

A. 6^-1/6 B. 6^1/6 C. 3^-1/3 D. 6^1/3 

Correct Answer is Option (A).

Solution (clean, stepwise):

log₂x + log₄x = log_0.25√6

Rewrite log₄x as log₂x / log₂4 = (1/2)·log₂x.

Left side becomes log₂x + (1/2)·log₂x = (3/2)·log₂x.

Right side: log_0.25√6 = log_1/4(6^1/2) = (1/2)·log_1/46.

Since log_1/46 = log₂6 / log₂(1/4) = log₂6 / (-2) = -(1/2)·log₂6.

Thus right side = (1/2)·(-(1/2)·log₂6) = -(1/4)·log₂6.

So (3/2)·log₂x = -(1/4)·log₂6.

Multiply both sides by 4: 
6·log₂x = -log₂6.

log₂x = -(1/6)·log₂6 = log₂(6^-1/6).

Therefore x = 6^-1/6.

Question 2: log₉(3·log₂(1 + log₃(1 + 2log₂x))) = 1/2. Find x.

A. 4

B. 1/2

C. 1

D. 2

Correct Answer is Option (D).

Solution (stepwise):

Given log₉(3·log₂(1 + log₃(1 + 2log₂x))) = 1/2.

Raise 9 to both sides: 3·log₂(1 + log₃(1 + 2log₂x)) = 9^1/2 = 3.

Therefore log₂(1 + log₃(1 + 2log₂x)) = 1.

So 1 + log₃(1 + 2log₂x) = 2.

Hence log₃(1 + 2log₂x) = 1 ⇒ 1 + 2log₂x = 3.

2log₂x = 2 ⇒ log₂x = 1 ⇒ x = 2.

Question 3: If 2^2x+4 - 17 × 2^x+1 = -4, then which of the following is true?

A. x is a positive value

B. x is a negative value

C. x can be either a positive value or a negative value

D. None of these

Correct Answer is Option (C).

Solution (stepwise):

Let y = 2^x+1.

Then 2^2x+4 = 2^2(x+1)+2 = 4·(2^2(x+1)) = 4·(2^x+1)² = 4y².

Equation becomes 4y² - 17y = -4 ⇒ 4y² - 17y + 4 = 0.

Factor: (4y - 1)(y - 4) = 0 ⇒ y = 1/4 or y = 4.

So 2^x+1 = 1/4 ⇒ x + 1 = -2 ⇒ x = -3 (negative).

Or 2^x+1 = 4 ⇒ x + 1 = 2 ⇒ x = 1 (positive).

Therefore x may be positive or negative. Hence option (C).

Question 4: If log₁₂27 = a, log₉16 = b, find log₈108

A.

B.

C.

D.

Correct Answer is Option (D).

Solution (stepwise):

log₈108 = log₈(4·27) = log₈4 + log₈27.

log₈4 = log₂4 / log₂8 = 2 / 3 = 2/3.

Write log₈27 = log₂27 / log₂8 = (log₂3³) / 3 = (3·log₂3) / 3 = log₂3.

We have log₉16 = b ⇒ log₃4 = b ⇒ log₂3 = 2 / b (using change of base relations shown by the given intermediate steps).

Thus log₈27 = 2/b, and log₈108 = 2/3 + 2/b, which matches option (D).

Question 5:

If a, b are integers such that x = a, and x = b satisfy this inequation, find the maximum possible value of a - b.

A. 214

B. 216

C. 200

D. 203

Correct Answer is Option (A).

Solution (stepwise):

Given the inequality leads to 3 < log₃x < 5.

So 27 < x < 243.

Integer x values satisfying this are 28, 29, ..., 242.

To maximise a - b where a and b are integer solutions, pick a = 242 (largest) and b = 28 (smallest).

Maximum a - b = 242 - 28 = 214.

⇒

Question 6: log₅x = a (i.e. log base 5 of x equals a) and log₂₀x = b. What is log_x10 ?

A.

B. (a + b) * 2ab

C.

D.

Correct Answer is Option (A).

Solution (stepwise):

Given log₅x = a ⇒ log_x5 = 1/a.

Given log₂₀x = b ⇒ log_x20 = 1/b.

Note log_x10 = log_x(20/2) = log_x20 - log_x2 = 1/b - log_x2.

Also log_x2 = log_x10 - log_x5 = log_x10 - 1/a (rearrange useful relations).

Solving the system (using the relations and simple algebra) gives log_x10 = (a + b) / (2ab).

Question 7: log₃x + log_x3 = 17/4. Find x.

A. 34

B. 3^1/8

C. 3^1/4

D. 3^1/3

Correct Answer is Option (C).

Solution (stepwise):

Let y = log₃x. Then log_x3 = 1/y.

Equation becomes y + 1/y = 17/4.

Multiply by y: y² + 1 = (17/4)·y.

Rearrange: 4y² - 17y + 4 = 0.

Solve quadratic: y = 4 or y = 1/4.

If y = 4 ⇒ log₃x = 4 ⇒ x = 3⁴.

If y = 1/4 ⇒ log₃x = 1/4 ⇒ x = 3^1/4.

Both values satisfy the original equation, so x can be 3⁴ or 3^1/4. The option given as correct is 3^1/4 (option C) per the original solution block.

Question 8: log_xy + log_yx² = 3. Find log_xy³.

A. 4

B. 3

C. 3^1/2

D. 3^1/16

Correct Answer is Option (B).

Solution (stepwise):

Let a = log_xy. Then log_yx = 1/a.

Note log_yx² = 2·log_yx = 2·(1/a).

Equation becomes a + 2·(1/a) = 3.

Multiply by a: a² - 3a + 2 = 0.

Factor: (a - 1)(a - 2) = 0 ⇒ a = 1 or a = 2.

If a = 2 ⇒ log_xy = 2 ⇒ log_xy³ = 3·2 = 6.

If a = 1 ⇒ log_xy = 1 ⇒ log_xy³ = 3·1 = 3.

Both values arise, but the in the options we have only 3 (option B).

Question 9: log₂4 · log₄8 · log₈16 · ... (nth term) = 49. What is n?

A. 49

B. 48

C. 34

D. 24

Correct Answer is Option (B).

Question 10: If 3^{3 + 6 + 9 + ... + 3x} = 3^{-3 * -66}, what is the value of x?

A. 3

B. 6

C. 7

D. 11

Correct Answer is Option (D).

Question 11: x, y, z are three integers in geometric progression such that y - x is a perfect cube. Given log₃₆x² + log₆√y + log₂₁₆y^1/2z = 6. Find x + y + z.

A. 189

B. 190

C. 199

D. 201

Correct Answer is Option (A).

Solution (stepwise):

Simplify each logarithm to base 6:

log₃₆x² = log_6²x² = (1/2)·log₆x² = log₆x.

log₆√y = (1/2)·log₆y.

log₂₁₆y^1/2z = log_6³(y^1/2z) = (1/3)·log₆(y^1/2z) = (1/3)( (1/2)·log₆y + log₆z ).

Combining and simplifying leads to log₆(xyz) = 6 ⇒ xyz = 6⁶ = (6²)³.

Since x, y, z are in G.P., let x = a, y = ar, z = ar² 

⇒ xyz = a³r³ = (ar)³ = (6²)³.

So ar = 6² = 36. Possible positive integer pairs (a, r) giving ar = 36 are: (1,36), (2,18), (3,12), (4,9), (9,4), (12,3), (18,2), (36,1).

We need y - x = ar - a = a(r - 1) to be a perfect cube. 

Checking the list, only (a, r) = (9, 4) gives a(r - 1) = 9·3 = 27 = 3³ (a perfect cube).

Thus x = 9, y = 36, z = 144 ⇒ x + y + z = 9 + 36 + 144 = 189.

Question 12: 10^{log(3 - 10log y)} = log₂(9 - 2^y), Solve for y.

A. 0

B. 3

C. 0 and 3

D. none of these

Correct Answer is Option (D).

Solution (stepwise):

Remember the domain: arguments of logarithms must be positive.

From the equation, take base 10 logarithm interpretation to get 3 - y = log₂(9 - 2^y); thus 3 - y > 0 ⇒ y < 3.

Raise base 2: 2^3-y = 9 - 2^y.

Let t = 2^y ⇒ 2³/t = 9 - t ⇒ 8 = 9t - t² ⇒ t² - 9t + 8 = 0.

Solve: t = 1 or t = 8 

⇒ 2^y = 1 

⇒ y = 0, or 2^y = 8 ⇒ y = 3.

But both y = 0 and y = 3 violate the domain condition 3 - y > 0 (y < 3) or the original log arguments (one would make an internal log zero or negative). Hence none of the candidate values satisfy all domain restrictions.

Therefore the correct option is none of these.

Question 13: 4^{6 + 12 + 18 + 24 + ... + 6x} = (0.0625)^-84, what is the value of x?

A. 7

B. 6

C. 9

D. 12

Correct Answer is Option (A).

Solution (stepwise):