CBSE Class 11 Mathematics Logarithms Competitive Exam MCQs Set 01

Answer: (b) \( \log |a| \)

Explanation:
1. For log ab to be defined, ab > 0, so a and b have the same sign.
2. Then \( \log ab = \log |ab| = \log |a| + \log |b| \).
3. Hence \( \log ab - \log |b| = \log |a| \).

Teacher's Note:
1. When a and b may be negative, write log ab = log |a| + log |b|; the plain rule log ab = log a + log b needs a, b > 0.
2. Keeping the modulus avoids taking the logarithm of a negative number.

Answer: (c) 2

Explanation:
1. \( \log_{0.5} 4 = \log_{2^{-1}} 2^2 = -2 \).
2. \( (\log_{0.5} 4)^2 = 4 \).
3. \( \sqrt{4} = 2 \) (the square root sign means the non-negative root).

Teacher's Note:
1. √(x²) = |x|, never a negative number, so −2 cannot be the value.
2. Writing 0.5 as 2⁻¹ turns the logarithm into a simple power comparison.

Answer: (b) 2

Explanation:
1. Using \( \log_a b \cdot \log_b c = \log_a c \) repeatedly, the product telescopes.
2. Product \( = \log_3 9 = \log_3 3^2 = 2 \).

Teacher's Note:
1. In a chain log_a b · log_b c · log_c d …, each base cancels the previous argument, leaving log of the last argument to the first base.
2. Writing each factor as log x / log y shows the cancellation clearly.

Answer: (c) \( 1 - 3\log_7 2 \)

Explanation:
1. \( \sqrt{7\sqrt{7\sqrt{7}}} = 7^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8}} = 7^{7/8} \).
2. \( \log_7 7^{7/8} = \frac{7}{8} \).
3. \( \log_7 \frac{7}{8} = \log_7 7 - \log_7 8 = 1 - 3\log_7 2 \).

Teacher's Note:
1. Nested square roots of the same base give exponents 1/2, 1/4, 1/8, … which add up easily.
2. Work from the inside outwards: first find the inner logarithm, then apply the outer one.

Answer: (d) 890

Explanation:
1. \( 81^{1/\log_5 3} = 81^{\log_3 5} = 3^{4\log_3 5} = 5^4 = 625 \).
2. \( 27^{\log_9 36} = 3^{3 \cdot \frac{1}{2}\log_3 36} = 3^{\log_3 6^3} = 216 \).
3. \( 3^{4/\log_7 9} = 3^{4\log_9 7} = 3^{2\log_3 7} = 7^2 = 49 \).
4. Sum = 625 + 216 + 49 = 890.

Teacher's Note:
1. Turn 1/log_a b into log_b a and bring every power to base 3 so that a^(log_a m) = m can be used.
2. Each term is a single number; add them only at the end.

Answer: (c) \( \log 2 \)

Explanation:
1. Write in prime factors: \( \frac{16}{15} = \frac{2^4}{3 \cdot 5},\ \frac{25}{24} = \frac{5^2}{2^3 \cdot 3},\ \frac{81}{80} = \frac{3^4}{2^4 \cdot 5} \).
2. Coefficient of log 2: 28 − 15 − 12 = 1; of log 3: −7 − 5 + 12 = 0; of log 5: −7 + 10 − 3 = 0.
3. Hence the expression \( = \log 2 \).

Teacher's Note:
1. Breaking every fraction into prime factors turns the problem into collecting coefficients of log 2, log 3 and log 5.
2. This method works for all such ‘weighted log’ sums.

Answer: (d) \( \frac{1}{2ab - 1} \)

Explanation:
1. \( ab = \log_4 5 \cdot \log_5 6 = \log_4 6 = \frac{\log_2 6}{2} \), so \( \log_2 6 = 2ab \).
2. \( \log_2 6 = 1 + \log_2 3 \Rightarrow \log_2 3 = 2ab - 1 \).
3. \( \log_3 2 = \frac{1}{\log_2 3} = \frac{1}{2ab - 1} \).

Teacher's Note:
1. Multiplying the given logs links the bases 4 and 6, which share the prime 2 with the target.
2. log₄ x = ½ log₂ x because 4 = 2².

Answer: (b) \( \frac{1}{5} \); (c) 5

Explanation:
1. \( \log_k x \cdot \log_5 k = \log_5 x \), so the equation is \( \log_5 x = \log_x 5 = \frac{1}{\log_5 x} \).
2. \( (\log_5 x)^2 = 1 \Rightarrow \log_5 x = \pm 1 \).
3. Hence x = 5 or \( x = \frac{1}{5} \); both options (b) and (c) are correct.

Teacher's Note:
1. An equation of the form t = 1/t gives t = ±1; do not forget the negative value.
2. log_a b · log_b c = log_a c is the key step here.

Answer: (c) 25

Explanation:
1. \( \log_5 a \cdot \log_a x = \log_5 x \).
2. \( \log_5 x = 2 \Rightarrow x = 5^2 = 25 \).

Teacher's Note:
1. The base a cancels in the chain, so a does not appear in the answer.
2. Change the log statement to exponential form to finish.

Answer: (c) \( \frac{1}{2}[\log a + \log b + 4\log 2] \)

Explanation:
1. Add 4ab to both sides: \( a^2 + 4ab + 4b^2 = 16ab \Rightarrow (a + 2b)^2 = 16ab \).
2. Taking log: \( 2\log(a + 2b) = \log 16 + \log a + \log b \).
3. \( \log(a + 2b) = \frac{1}{2}[\log a + \log b + 4\log 2] \).

Teacher's Note:
1. Complete the square to get the required expression a + 2b on one side.
2. log 16 = 4 log 2.

Answer: (c) 5

Explanation:
1. \( \log_4 256 = \log_4 4^4 = 4 \); \( \log_2 4 = 2 \); \( \log_2 2 = 1 \).
2. \( 2\log_{\sqrt{2}} 2 = 2 \times 2 = 4 \), since \( (\sqrt{2})^2 = 2 \).
3. A = 1 + 4 = 5.

Teacher's Note:
1. Evaluate nested logarithms from the inside out.
2. log over base √2 of 2 is 2 because (√2)² = 2.

Answer: (d) \( \frac{2y}{3} \)

Explanation:
1. \( \log_{1000} x^2 = \log_{10^3} x^2 = \frac{2}{3}\log_{10} x \).
2. \( = \frac{2y}{3} \).

Teacher's Note:
1. Use log over base a^β of n^α = (α/β) log_a n.
2. 1000 = 10³, so the base change is immediate.

Answer: (b) \( (1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1} \)

Explanation:
1. \( 1 + x = \log_a a + \log_a bc = \log_a (abc) \), so \( \frac{1}{1 + x} = \log_{abc} a \).
2. Similarly \( \frac{1}{1 + y} = \log_{abc} b,\ \frac{1}{1 + z} = \log_{abc} c \).
3. Sum \( = \log_{abc} (abc) = 1 \).

Teacher's Note:
1. Adding 1 = log_a a makes every expression a logarithm of the same number abc.
2. Taking reciprocals swaps base and argument, producing a common base abc.

Answer: (c) 2bc

Explanation:
1. \( abc = \log_{48} 12 \) (chain rule), so \( 1 + abc = \log_{48} 48 + \log_{48} 12 = \log_{48} 576 = \log_{48} 24^2 = 2\log_{48} 24 \).
2. \( bc = \log_{36} 24 \cdot \log_{48} 36 = \log_{48} 24 \).
3. Hence 1 + abc = 2bc.

Teacher's Note:
1. Products of logarithms with matching base and argument collapse into a single logarithm.
2. 576 = 24² is the key observation.

Answer: (b) 1

Explanation:
1. \( x = \log_a b,\ y = \log_b c,\ z = \log_c a \).
2. \( xyz = \log_a b \cdot \log_b c \cdot \log_c a = \log_a a = 1 \).

Teacher's Note:
1. Converting each power statement to logarithmic form gives a cyclic chain.
2. Alternatively, a = c^z = b^(yz) = a^(xyz), so xyz = 1.

Answer: (b) \( x^x y^y z^z = 1 \)

Explanation:
1. Let \( \log x = k(y - z),\ \log y = k(z - x),\ \log z = k(x - y) \).
2. \( \log(x^x y^y z^z) = x\log x + y\log y + z\log z = k[x(y - z) + y(z - x) + z(x - y)] = 0 \).
3. Hence \( x^x y^y z^z = 1 \).

Teacher's Note:
1. Put each logarithm equal to k times its ratio term, then test the options by taking logs.
2. Expressions like x(y − z) + y(z − x) + z(x − y) always sum to zero.

Answer: (d) \( 1 - \ln 2 \)

Explanation:
1. \( \log_{2^n} 2 = \frac{1}{n} \), so the series is \( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \)
2. We know \( \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \)
3. So the series \( = 1 - \ln 2 \).

Teacher's Note:
1. Every term has a base that is a power of 2, so each term is simply 1/n.
2. The standard series ln 2 = 1 − 1/2 + 1/3 − … is worth remembering.

Answer: (c) 9

Explanation:
1. \( \log(3^{12} \times 2^8) = 12 \times 0.47712 + 8 \times 0.30103 = 5.72544 + 2.40824 = 8.13368 \).
2. The characteristic is 8, so the number of digits is 8 + 1 = 9.

Teacher's Note:
1. Number of digits of N = (characteristic of log₁₀ N) + 1.
2. Check: 3¹² × 2⁸ = 136048896 has 9 digits.

Answer: (d) 0

Explanation:
1. Pair r° with (90 − r)°: \( \tan r^\circ \cdot \tan(90 - r)^\circ = \tan r^\circ \cot r^\circ = 1 \).
2. So the sum \( = \log_3 (\tan 1^\circ \tan 2^\circ \cdots \tan 89^\circ) = \log_3 1 = 0 \) (the middle term tan 45° = 1).

Teacher's Note:
1. Sum of logs = log of the product; complementary angles make the product 1.
2. tan 45° = 1 is the unpaired middle term.

Answer: (a) \( \frac{n(n + 1)}{2}\log_a 2 \)

Explanation:
1. \( \frac{1}{\log_{2^n} a} = \log_a 2^n = n\log_a 2 \).
2. Sum \( = (1 + 2 + \ldots + n)\log_a 2 = \frac{n(n + 1)}{2}\log_a 2 \).

Teacher's Note:
1. 1/log_b a = log_a b moves the variable power into the argument, where it comes down as a multiplier.
2. Do not mix up log_a 2 and log₂ a in the options.

Answer: (d) \( e^x < 1 - x \) for x > 0

Explanation:
1. For x > 0, \( e^x = 1 + x + \frac{x^2}{2!} + \ldots > 1 + x \), so (c) is true and (a) follows by taking logs.
2. (b) is true since \( \log(1 + x) = -\log\left(1 - \frac{x}{1 + x}\right) > \frac{x}{1 + x} \).
3. (d) is false: for x > 0, \( e^x > 1 > 1 - x \).

Teacher's Note:
1. For x > 0, eˣ is always greater than 1 while 1 − x is less than 1.
2. The inequalities x/(1 + x) < log(1 + x) < x hold for all x > 0.

Answer: (c) x = 4

Explanation:
1. \( \log_7 \log_5(\ldots) = 0 \Rightarrow \log_5 \sqrt{x^2 + 5 + x} = 1 \Rightarrow \sqrt{x^2 + x + 5} = 5 \).
2. \( x^2 + x + 5 = 25 \Rightarrow x^2 + x - 20 = 0 \Rightarrow (x + 5)(x - 4) = 0 \).
3. x = 4 or x = −5; among the options, x = 4.

Teacher's Note:
1. Peel off logarithms one at a time using log_a t = 0 ⇒ t = 1 and log_a t = 1 ⇒ t = a.
2. Always check that the values found keep every logarithm defined.

Answer: (b) An irrational number

Explanation:
1. \( \log_4 18 = \frac{\log_2 18}{2} = \frac{1 + 2\log_2 3}{2} \).
2. If it were rational, \( \log_2 3 = \frac{p}{q} \) would give \( 2^p = 3^q \), impossible (even ≠ odd).
3. Hence log₄ 18 is irrational.

Teacher's Note:
1. To prove a logarithm is irrational, assume it equals p/q and reach an even = odd contradiction.
2. Here everything reduces to the irrationality of log₂ 3.

Answer: (a) 81

Explanation:
1. \( 0.1 + 0.01 + \ldots = \frac{0.1}{1 - 0.1} = \frac{1}{9} \).
2. \( \log_{\sqrt{20}} \frac{1}{9} = 2\log_{20} 9^{-1} = -2\log_{20} 9 \).
3. \( (0.05)^{-2\log_{20} 9} = \left(\frac{1}{20}\right)^{-2\log_{20} 9} = 20^{\log_{20} 81} = 81 \).

Teacher's Note:
1. Sum the infinite G.P. first; then 0.05 = 1/20 matches the base √20.
2. a^(log_a m) = m finishes the problem.

Answer: (a) 1

Explanation:
1. Let A = log a, B = log b, C = log c. The condition is \( \frac{A^2}{BC} + \frac{B^2}{CA} + \frac{C^2}{AB} = 3 \).
2. So \( A^3 + B^3 + C^3 - 3ABC = 0 \Rightarrow (A + B + C)(A^2 + B^2 + C^2 - AB - BC - CA) = 0 \).
3. The second factor is zero only if A = B = C, which is not possible (a, b, c distinct). So A + B + C = 0.
4. log(abc) = 0 ⇒ abc = 1.

Teacher's Note:
1. Change every logarithm to a common base so that the condition becomes an algebraic identity.
2. x³ + y³ + z³ = 3xyz gives x + y + z = 0 or x = y = z.

Answer: (c) \( 4 \cdot \frac{3 - a}{3 + a} \)

Explanation:
1. Let t = log₂ 3. Then \( a = \frac{3t}{2 + t} \Rightarrow t = \frac{2a}{3 - a} \).
2. \( \log_6 16 = \frac{4}{1 + t} = \frac{4}{1 + \frac{2a}{3 - a}} = \frac{4(3 - a)}{3 + a} \).

Teacher's Note:
1. Express both logarithms in terms of the single unknown log₂ 3.
2. 12 = 2²·3 and 6 = 2·3, so base 2 is the natural choice.

Answer: (c) 1

Explanation:
1. \( \frac{1}{\log_k n} = \log_n k \).
2. Sum \( = \log_n (2 \cdot 3 \cdot 4 \cdots 1983) = \log_n 1983! = \log_n n = 1 \).

Teacher's Note:
1. Reciprocal of log_k n is log_n k, and sums of logs with the same base multiply the arguments.
2. Including the factor 1 does not change 1983!, so the product is exactly n.

Answer: (a) xyz = 1; (b) \( x^a y^b z^c = 1 \); (c) \( x^{b + c} y^{c + a} z^{a + b} = 1 \); (d) \( xyz = x^a y^b z^c \)

Explanation:
1. Let each ratio be k: \( \log x = k(b - c),\ \log y = k(c - a),\ \log z = k(a - b) \).
2. Σ log x = k·0 = 0 ⇒ xyz = 1; Σ a log x = k[a(b − c) + b(c − a) + c(a − b)] = 0 ⇒ xᵃyᵇzᶜ = 1.
3. Σ (b + c) log x = k[(b² − c²) + (c² − a²) + (a² − b²)] = 0 ⇒ \( x^{b + c} y^{c + a} z^{a + b} = 1 \).
4. Since xyz and xᵃyᵇzᶜ are both 1, (d) is also true. All four options are correct.

Teacher's Note:
1. Setting the equal ratios to k and testing each option by taking logarithms is the standard approach.
2. Cyclic sums like a(b − c) + b(c − a) + c(a − b) are always zero.

Answer: (b) 1

Explanation:
1. \( \log_{x_n} x_n^{x_{n-1}^{\cdots^{x_1}}} = x_{n-1}^{\cdots^{x_1}} \).
2. Then \( \log_{x_{n-1}} \) of this gives \( x_{n-2}^{\cdots^{x_1}} \), and so on.
3. Finally \( \log_{x_1} x_1 = 1 \).

Teacher's Note:
1. Each logarithm removes the lowest base of the power tower: log_a a^t = t.
2. Work from the innermost logarithm outwards.

Answer: (d) None of these

Explanation:
1. log₂(x + 5) is increasing and 6 − x is decreasing, so their graphs meet at most once.
2. x = 3 gives log₂ 8 = 3 = 6 − 3, so there is exactly one solution.
3. Since 1 is not among (a)–(c), the answer is ‘None of these’.

Teacher's Note:
1. An increasing function and a decreasing function can intersect at most once.
2. Try small integer values to find the actual solution quickly.

Answer: (b) 1

Explanation:
1. Put t = log₁₆ x: \( t^2 - t + \log_{16} k = 0 \). Each real t gives exactly one x > 0.
2. Exactly one solution ⇒ equal roots ⇒ \( 1 - 4\log_{16} k = 0 \Rightarrow \log_{16} k = \frac{1}{4} \Rightarrow k = 2 \).
3. So there is only one such value of k.

Teacher's Note:
1. Substituting t = log x turns the equation into a quadratic; x and t correspond one-to-one.
2. A single solution needs discriminant zero.

Answer: (a) One positive integral value; (b) One irrational value; (c) Two positive rational values

Explanation:
1. Take log₃ and put t = log₃ x: \( t\left[\frac{3}{4}t^2 + t - \frac{5}{4}\right] = \frac{1}{2} \Rightarrow 3t^3 + 4t^2 - 5t - 2 = 0 \).
2. \( (t - 1)(3t + 1)(t + 2) = 0 \Rightarrow t = 1, -\frac{1}{3}, -2 \).
3. x = 3, \( 3^{-1/3} \), \( \frac{1}{9} \): one positive integer (3), one irrational value (\( 3^{-1/3} \)) and two positive rationals (3 and 1/9).

Teacher's Note:
1. Taking logarithm to base 3 of both sides converts an exponential equation in x into a polynomial in log₃ x.
2. Try t = 1 first; it often turns out to be a root.

Answer: (a) x > y

Explanation:
1. 5⁴ = 625 < 1000, so x = log₅ 1000 > 4.
2. 7⁴ = 2401 > 2058, so y = log₇ 2058 < 4.
3. Hence x > y.

Teacher's Note:
1. Compare both logarithms with a convenient integer between them.
2. Powers of the base near the argument give quick bounds.

Answer: (b) \( \left(\frac{1}{3}, \frac{1}{2}\right) \)

Explanation:
1. \( 3^2 = 9 < 20 \Rightarrow \log_{20} 3 < \frac{1}{2} \).
2. \( 3^3 = 27 > 20 \Rightarrow \log_{20} 3 > \frac{1}{3} \).
3. So \( \frac{1}{3} < \log_{20} 3 < \frac{1}{2} \).

Teacher's Note:
1. log_b a > 1/n exactly when aⁿ > b (for a, b > 1).
2. Comparing integer powers avoids any decimal calculation.

Answer: (a) 2

Explanation:
1. LHS \( = \log_\pi 3 + \log_\pi 4 = \log_\pi 12 \).
2. π² ≈ 9.87 < 12 < π³, so \( 2 < \log_\pi 12 < 3 \).
3. Hence x = 2 satisfies the inequality; 3, 3.5 and π do not.

Teacher's Note:
1. Reciprocals of logarithms swap base and argument, after which the logs can be added.
2. Compare 12 with powers of π to bound the value.

Answer: (a) 4

Explanation:
1. Base 1/√2 lies between 0 and 1, so the inequality means 0 < sin x < 1.
2. Multiples of π/4 in [0, 4π] with 0 < sin x < 1: π/4, 3π/4, 9π/4, 11π/4.
3. So there are 4 values.

Teacher's Note:
1. With a base between 0 and 1, log_a t > 0 means 0 < t < 1.
2. Exclude points where sin x = 1 (π/2, 5π/2) and where sin x ≤ 0.

Answer: (b) [2, 4]

Explanation:
1. x² − 6x + 12 = (x − 3)² + 3 > 0 always, so the log is defined for all x.
2. Base 1/2 < 1, so the inequality reverses: \( x^2 - 6x + 12 \le \left(\frac{1}{2}\right)^{-2} = 4 \).
3. x² − 6x + 8 ≤ 0 ⇒ (x − 2)(x − 4) ≤ 0 ⇒ x ∈ [2, 4].

Teacher's Note:
1. When the base is between 0 and 1, removing the logarithm reverses the inequality sign.
2. Always check the domain condition (argument > 0) as well.

Answer: (b) (4, +∞)

Explanation:
1. Domain: x > 1. \( 2^{\log_{\sqrt{2}} (x - 1)} = 2^{2\log_2 (x - 1)} = (x - 1)^2 \).
2. (x − 1)² > x + 5 ⇒ x² − 3x − 4 > 0 ⇒ (x − 4)(x + 1) > 0 ⇒ x < −1 or x > 4.
3. Combining with x > 1 gives x ∈ (4, ∞).

Teacher's Note:
1. log over base √2 equals 2 log₂, which makes the power simplify to (x − 1)².
2. Do not forget the domain x > 1, which removes x < −1.

Answer: (c) \( [-1, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, 3] \)

Explanation:
1. Need \( 0 < x^2 - 2x - 2 \le 1 \).
2. x² − 2x − 3 ≤ 0 ⇒ −1 ≤ x ≤ 3; x² − 2x − 2 > 0 ⇒ \( x < 1 - \sqrt{3} \) or \( x > 1 + \sqrt{3} \).
3. Intersection: \( [-1, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, 3] \).

Teacher's Note:
1. log₁₀ t ≤ 0 means 0 < t ≤ 1, not just t ≤ 1.
2. The end points 1 ± √3 are excluded because the log is undefined there.