Logarithm Aptitude Questions and Answers (Solved MCQs)

c) 6 — We have log₂32 = 5, log₄64 = 3 and log₈64 = 2. Therefore, the required value is 5 + 3 − 2 = 6.

a) 3 — From log_a81 = 4, we get a⁴ = 81. Since a must be positive, a = 3.

b) √13 — The domain requires x−2 > 0 and x+2 > 0, so x > 2. Combining the logarithms gives log₃[(x−2)(x+2)] = 2. Therefore, x²−4 = 9, giving x² = 13. Of the two algebraic roots, only x = √13 satisfies x > 2.

c) 4 — Using log_ab × log_bc = log_ac, the expression becomes log₂16 = 4.

c) 1.9030 — Since 80 = 2³×10, log₁₀80 = 3log₁₀2 + log₁₀10 = 3(0.3010)+1 = 1.9030.

b) −3 — Let log_1/327 = y. Then (1/3)^y = 27. Writing both sides as powers of 3 gives 3^−y = 3³. Hence, −y = 3 and y = −3.

c) 8 — From log_x16 = 4/3, we get x^4/3 = 16. Raising both sides to the power 3/4 gives x = 16^3/4 = (2⁴)^3/4 = 2³ = 8.

b) 2 — First, log₃81 = 4 because 3⁴ = 81. Therefore, the expression becomes log₂4 = 2.

b) 1 — By the change-of-base formula, the expression is (log b/log a)×(log c/log b)×(log a/log c). All the terms cancel, leaving 1.

b) 4 — Using the quotient law, log₂12 − log₂3 = log₂(12/3) = log₂4 = 2. Also, log₄16 = 2. Hence, the total is 2+2 = 4.

b) 3p+2q — Since 72 = 2³×3², log₁₀72 = log₁₀(2³×3²) = 3log₁₀2 + 2log₁₀3 = 3p+2q.

c) 3/2 — Since 1/8 = 2⁻³, log₂(1/8) = −3. Also, 1/16 = 4⁻², so log₄(1/16) = −2. Therefore, the required value is (−3)/(−2) = 3/2.

b) 3 — Applying the identity log_ab × log_bc = log_ac repeatedly, the expression becomes log₂8 = 3.

c) (1,2) ∪ (3,∞) — The base must satisfy x−1 > 0 and x−1 ≠ 1, giving x > 1 and x ≠ 2. The argument must be positive: x²−5x+6 = (x−2)(x−3) > 0, which gives x < 2 or x > 3. Combining the conditions produces (1,2) ∪ (3,∞).

c) 8/3 — The numerator simplifies to log₂16 = 4. Also, log₄8 = log 8/log 4 = 3/2. Therefore, the required value is 4 ÷ (3/2) = 4×(2/3) = 8/3.

b) 5 — The domain requires x > 3. Combining the logarithms gives log₂[(x−1)(x−3)] = 3. Therefore, (x−1)(x−3) = 8, giving x²−4x−5 = 0. Thus, x = 5 or x = −1. Only x = 5 satisfies the domain.

c) 7/2 — The domain requires x > 2. Using the quotient law gives log₃[(x+1)/(x−2)] = 1. Therefore, (x+1)/(x−2) = 3. Solving x+1 = 3x−6 gives 2x = 7, so x = 7/2.

b) 8 — From log_x64 = 2, we get x² = 64. Algebraically, x = ±8, but a logarithmic base must be positive and cannot equal 1. Therefore, x = 8.

c) 16 — Let log₂x = t. Since log₄x = log₂x/log₂4 = t/2, the equation becomes t+t/2 = 6. Thus, 3t/2 = 6 and t = 4. Hence, x = 2⁴ = 16.

c) 512 — From log₃(log₂x) = 2, we obtain log₂x = 3² = 9. Therefore, x = 2⁹ = 512.

c) 6 — The equation gives x²−6x+13 = 5. Therefore, x²−6x+8 = 0, which factors as (x−2)(x−4) = 0. The solutions are 2 and 4, and both make the logarithmic argument equal to 5. Hence, α+β = 2+4 = 6.

c) 5 — The domain requires x > 2. Combining the logarithms gives log₂[x(x−2)] = log₂(3x). Since the bases are equal, x(x−2) = 3x. As x > 2, division by x is valid, giving x−2 = 3. Therefore, x = 5.

c) 5 — The equation implies (x−1)² = 16. However, the base x−1 must be positive and cannot equal 1. Therefore, x−1 = 4 rather than −4, giving x = 5.

c) 9 — The domain requires x > 2. Using logarithmic laws gives log₃[x²/(x−2)] = 2. Therefore, x²/(x−2) = 9, so x²−9x+18 = 0. Factoring gives (x−3)(x−6) = 0. Both 3 and 6 satisfy the domain, so α+β = 3+6 = 9.

b) 8 — Let log₂x = t. Then log₄(8x) = log₂(8x)/log₂4 = (3+t)/2. Hence, t = (3+t)/2, giving 2t = 3+t and t = 3. Therefore, x = 2³ = 8.

b) 4+√17 — The domain requires x > 1. Combining the logarithms gives (x−1)(x+1) = 8x. Thus, x²−1 = 8x, or x²−8x−1 = 0. Solving gives x = 4±√17. Since 4−√17 is negative and violates x > 1, the valid solution is x = 4+√17.

c) 262,144 — First, log₃(log₄x) = 2¹ = 2. Therefore, log₄x = 3² = 9. Hence, x = 4⁹ = 262,144.

b) 2 — Combining the logarithms gives log_x(2×8) = log_x16 = 4. Therefore, x⁴ = 16. Since a logarithmic base must be positive, x = 2.

b) 5 — The right side requires x > 2, while the left side requires (x−2)(x−3) > 0. Together, these conditions give x > 3. Since 1 = log₂2, the equation becomes log₂[(x−2)(x−3)] = log₂[2(x−2)]. Therefore, (x−2)(x−3) = 2(x−2). Since x > 3, x−2 ≠ 0, so x−3 = 2 and x = 5.

b) 10 — Adding the two equations gives 2log₂x = 6, so log₂x = 3 and x = 8. Subtracting the second equation from the first gives 2log₂y = 4, so log₂y = 2 and y = 4. Therefore, x+y = 8+4 = 12. Hence, the correct answer is c) 12.

b) 5/3 — Using the change-of-base formula with base 2, log₈32 = log₂32/log₂8 = 5/3.

b) (2+a)/(1+a) — Using the change-of-base formula, log₆12 = log₂12/log₂6. Now, log₂12 = log₂(4×3) = 2+a, while log₂6 = log₂(2×3) = 1+a. Therefore, log₆12 = (2+a)/(1+a).

c) 3/2 — By the change-of-base formula, log₃5 = log 5/log 3 and log₂₅27 = log 27/log 25 = 3log 3/(2log 5). Their product is (log 5/log 3)×(3log 3/2log 5) = 3/2.

b) 6 — Using log_ab × log_bc = log_ac, we get log_ac = 2×3 = 6.

b) 16 — The number of digits in a positive integer N is ⌊log₁₀N⌋+1. Here, log₁₀(2⁵⁰) = 50×0.3010 = 15.05. Therefore, the number of digits is ⌊15.05⌋+1 = 15+1 = 16.

b) −3 — Since 0.00472 = 4.72×10⁻³, log₁₀(0.00472) = log₁₀4.72−3. Therefore, its characteristic is −3, conventionally written as bar 3.

b) 14 — Since log₁₀5 = log₁₀(10/2) = 1−0.3010 = 0.6990, we get log₁₀(5²⁰) = 20×0.6990 = 13.98. Hence, the number of digits is ⌊13.98⌋+1 = 14.

b) 5²⁸ — Taking common logarithms, log₁₀(3⁴⁰) = 40×0.4771 = 19.084, whereas log₁₀(5²⁸) = 28×0.6990 = 19.572. Since 19.572 > 19.084, we have 5²⁸ > 3⁴⁰.

b) 4.495 — We have pH = −log₁₀(3.2×10⁻⁵) = 5−log₁₀3.2. Since 3.2 = 32/10, log₁₀3.2 = 5log₁₀2−1 = 5(0.3010)−1 = 0.5050. Therefore, pH = 5−0.5050 = 4.495.

c) 100 times — The ratio of the intensities is 10⁷/10⁵ = 10² = 100. Therefore, the magnitude-7 earthquake is 100 times as intense according to the given model.

b) 9.01 years — If the investment doubles in t years, then (1.08)^t = 2. Taking logarithms gives tlog₁₀1.08 = log₁₀2. Thus, t = 0.3010/0.0334 ≈ 9.01 years. If interest is credited only at the end of complete years, the balance will first exceed twice the principal after 10 years.

b) 1 — Using base 2, log₄49 = log₂49/log₂4 = 2log₂7/2 = log₂7. Therefore, log₂7 ÷ log₄49 = 1.

c) log₂27 — Let log₂3 = a. Then log₄9 = log 3²/log 2² = a, and log₈27 = log 3³/log 2³ = a. Therefore, the sum is 3a = 3log₂3 = log₂(3³) = log₂27.

c) 300 — The antilogarithm of 2.4771 is 10^2.4771. Since 2.4771 = 2+0.4771 and 10^0.4771 = 3, we get 10²×3 = 300.

b) 85 — We have log₁₀(7¹⁰⁰) = 100×0.8451 = 84.51. Therefore, the number of digits in 7¹⁰⁰ is ⌊84.51⌋+1 = 84+1 = 85.

b) 1 < x < 5 — The logarithmic argument requires x−1 > 0, so x > 1. Since the base 1/2 lies between 0 and 1, the logarithmic function is decreasing. Therefore, log_1/2(x−1) > −2 implies x−1 < (1/2)⁻² = 4. Thus, x < 5, and the solution is 1 < x < 5.

b) 64 — Let log₂x = t. Then log₄x = t/2 and log₈x = t/3. Therefore, t×(t/2)×(t/3) = 36, giving t³/6 = 36. Hence, t³ = 216 and t = 6. Thus, x = 2⁶ = 64.

b) 1 — Let t = log₄x. Using the change-of-base formula, log_x16 = log₄16/log₄x = 2/t. Thus, 2/t = t, giving t² = 2 and t = ±√2. Therefore, the solutions are α = 4^√2 and β = 4^−√2. Their product is 4^√2−√2 = 1.

c) 10 — The domain requires 1 < x < 9. Combining the logarithms gives (x−1)(9−x) = 8. Expanding and simplifying gives x²−10x+17 = 0. By the sum-of-roots formula, α+β = 10. Both roots, 5±2√2, lie in the required domain.

b) 729 — Let log₃x = t. Then log₉x = t/2 and log₂₇x = t/3. Hence, t+t/2+t/3 = 11. This gives 11t/6 = 11, so t = 6. Therefore, x = 3⁶ = 729.

b) [1,2) ∪ (3,4] — The logarithmic argument must be positive: (x−2)(x−3) > 0, giving x < 2 or x > 3. Since the base 2 is greater than 1, the inequality is equivalent to 0 < x²−5x+6 ≤ 2. The upper inequality gives x²−5x+4 ≤ 0, or (x−1)(x−4) ≤ 0, so 1 ≤ x ≤ 4. Intersecting the two conditions gives [1,2) ∪ (3,4].

c) 4√2 — Let t = log₂x. Then log_x2 = 1/t. Thus, t+1/t = 5/2. Multiplying by 2t gives 2t²−5t+2 = 0, so (2t−1)(t−2) = 0. Hence, t = 1/2 or t = 2. The corresponding values of x are √2 and 4. Therefore, αβ = 4√2.

b) √13 — The domain requires x > 2. Since 2 = log₃9, the equation becomes log₃(x+2)+log₃(x−2) = log₃9. Therefore, (x+2)(x−2) = 9, giving x²−4 = 9 and x² = 13. Only x = √13 satisfies x > 2.

b) 4 — From log_x16 = 4, we get x⁴ = 16. Since x must be positive, x = 2. Next, log_y2 = 1/2 implies y^1/2 = 2. Therefore, y = 4.

c) 4,096 — Let the common logarithmic value be k. Then x = 2^k, y = 4^k = 2^2k and z = 8^k = 2^3k. Therefore, xyz = 2^6k = 2²⁴, giving 6k = 24 and k = 4. Hence, z = 2¹² = 4,096.

c) 11 — The equation is equivalent to x²−6x+k = 2, or x²−6x+(k−2) = 0. For exactly one real solution, its discriminant must be zero. Therefore, 36−4(k−2) = 0, giving 44−4k = 0 and k = 11. The repeated solution is x = 3, for which the logarithmic argument equals 2 and is valid.

d) log₂25 — The domain is −1 < x < 9. Combining the logarithms gives log₂[(x+1)(9−x)]. Now, (x+1)+(9−x) = 10. For two positive numbers with a fixed sum of 10, their product is maximum when both equal 5. This occurs when x+1 = 9−x, giving x = 4. The maximum product is 25, so the maximum logarithmic value is log₂25.

b) 103 — For 3ⁿ to contain 50 digits, we need 49 ≤ log₁₀(3ⁿ) < 50. Thus, 49 ≤ 0.4771n < 50. Dividing by 0.4771 gives approximately 102.70 ≤ n < 104.80. Therefore, the smallest integer satisfying the condition is n = 103.

b) 4 — The domain requires x > 1. Since log₂(x−1) = 2log₄(x−1), the equation becomes log₄[(x−1)²] = log₄(x+5). Therefore, (x−1)² = x+5. Simplifying gives x²−3x−4 = 0, or (x−4)(x+1) = 0. The roots are 4 and −1, but only x = 4 satisfies the domain.

b) 16 — Let y = log₂x. The logarithmic arguments require y > 2. Combining the logarithms gives log₂[y(y−2)] = 3, so y(y−2) = 8. Therefore, y²−2y−8 = 0, which factors as (y−4)(y+2) = 0. Since y > 2, y = 4. Hence, log₂x = 4 and x = 2⁴ = 16.