NDA Logarithms previous year questions with solutions

Answer: (a) bd²f³

Method. convert each logarithm to exponential form

log_b a = p gives a = b^p; log_d c = 2p gives c = d^2p; log_f e = 3p gives e = f^3p.
ace = b^p d^2p f^3p = (b d² f³)^p.
(ace)^1/⁠p = b d² f³.

Answer: (a) 0

Method. reduce each term with log₅ 2¹⁰ and log₅ 5⁵

110 log₅ 1024 = 110 log₅ 2¹⁰ = log₅ 2.
15 log₅ 3125 = 15 log₅ 5⁵ = 1.
Expression = log₅ 2 − log₅ 10 + 1 = log₅ 210 + 1 = log₅ 5^-1 + 1 = -1 + 1 = 0.

Answer: (d) 4

Method. log laws: 4 log 5 + 4 log 2 = 4 log 10

f(4) = log₁₀ 5 and f(1) = log₁₀ 2.
4f(4) + 5f(1) − log₁₀ 2 = 4 log₁₀ 5 + 5 log₁₀ 2 − log₁₀ 2 = 4(log₁₀ 5 + log₁₀ 2).
= 4 log₁₀ 10 = 4.

Answer: (c) 1 − 3 log₇ 2

Method. write the nested root as a single power of 7

⁠√(7√(7√7))⁠ = 7^1/⁠2 · 7^1/⁠4 · 7^1/⁠8 = 7^7/⁠8.
log₇ 7^7/⁠8 = 78.
log₇ 78 = log₇ 7 − log₇ 8 = 1 − 3 log₇ 2.

Answer: (b) 196

Method. change of base to the common prime: log_a^m aⁿ = n/⁠m

log₉ 27 = log_3² 3³ = 32.
log₈ 32 = log_2³ 2⁵ = 53.
Sum = 32 + 53 = 9 + 106 = 196.

Answer: (c) -0.4

Method. take log₁₀ of both sides; log₁₀ 0.2 = log₁₀ 2 − 1

x log₁₀ 0.2 = log₁₀ 2.
log₁₀ 0.2 = log₁₀ 210 = 0.3010 − 1 = -0.6990.
x = 0.3010-0.6990 = -0.4306…, which is -0.4 to the nearest tenth.

Answer: (b) 1

Method. complete the square; log₁₀ is increasing

x² + 2x + 11 = (x + 1)² + 10 ≥ 10, with equality at x = -1.
log₁₀ is increasing, so f(x) ≥ log₁₀ 10 = 1.
Minimum value 1, at x = -1.

Answer: (d) x > 0 excluding x = 1

Method. the base of a logarithm must be positive and not 1

log_x 10 = log 10log x is defined when the base x > 0 and x ≠ 1 (the argument 10 is already positive).
At x = 1, log x = 0 and the quotient is undefined.
Domain: x > 0, x ≠ 1.

Answer: (b) Negative power of 10 is between 0 and 1

Method. log₁₀ a = y means a = 10^y; the values of 10^y lying in (0, 1) are exactly the negative powers

log₁₀ a = y means 10^y = a.
For y ≥ 0, 10^y ≥ 1; for y < 0, 10^y = 110^-y lies strictly between 0 and 1.
So the numbers between 0 and 1 are precisely the negative powers of 10, and 0 < a < 1 forces y = log₁₀ a < 0.
The statement that justifies this is 'a negative power of 10 is between 0 and 1'; (a) and (c) state only half of that range, and (d) is false.

Options (a) and (c) are also true statements; each is half of (b). Where more than one option is true, the strongest, the one that implies the others, is the answer: (b).

Answer: (d) 1

Method. k = 2 − (t + 1/⁠t) with t = log_x y, and t + 1/⁠t ≥ 2 for t > 0

k = (1 − log_x y) + (1 − log_y x) = 2 − (t + 1t), where t = log_x y.
For x ≥ y > 1, 0 < t ≤ 1, so t + 1t ≥ 2 and k ≤ 0 (k = 0 when x = y).
k = -1 needs t + 1t = 3, t = 3 - ⁠√52⁠ ∈ (0, 1); k = -12 needs t = 12: both possible.
k = 1 would need t + 1t = 1 < 2, impossible.

Answer: (b) 1

Method. log_a b · log_b a = 1 and log₁₀ 10^x = x

log₁₀ 2 · log₂ 10 = 1.
log₁₀(10^x) = x.
So 1 + x = 2, and x = 1.

Answer: (b) 1

Method. 1/⁠log_k n = log_n k, then add the logarithms

1log_k n = log_n k for k = 2, 3, …, 100 (n = 100! > 1).
Sum = log_n(2 · 3 · … · 100) = log_n(100!) = log_n n = 1.

Answer: (c) (1 + x)^-1 + (1 + y)^-1 + (1 + z)^-1 = 1

Method. 1 + log_c(ab) = log_c(abc), so each reciprocal is log_abc of one letter

1 + x = log_c c + log_c(ab) = log_c(abc), so (1 + x)^-1 = log_abc c.
Likewise (1 + y)^-1 = log_abc a and (1 + z)^-1 = log_abc b.
Sum = log_abc(abc) = 1 (a, b, c positive, none equal to 1, abc ≠ 1).

Answer: (b) 1

Method. 1/⁠log_k n = log_n k, then add the logarithms

1log_k n = log_n k for each k = 2, 3, …, 2017 (n = 2017! > 1, so n ≠ 1).
Sum = log_n 2 + log_n 3 + … + log_n 2017 = log_n(2 · 3 · … · 2017) = log_n(2017!).
Since n = 2017!, the sum is log_n n = 1.

Answer: (a) 1log_100! N

Method. 1/⁠log_k N = log_N k, then add the logarithms

1log_k N = log_N k (N > 0, N ≠ 1), for k = 2, 3, …, 100.
Sum = log_N(2 · 3 · … · 100) = log_N(100!).
log_N(100!) = 1log_100! N.

Answer: (a) log_e 9

Method. 1/⁠log_b a = log_a b, then an infinite GP

1log₃ e^m = 1m log₃ e = log_e 3m.
The exponents are m = 1, 2, 4, 8, …, so the sum is log_e 3 (1 + 12 + 14 + …).
The GP has sum 11 − 12 = 2, so the total is 2 log_e 3 = log_e 9.

Answer: (d) 0

Method. 1 − log₁₀ 2 = log₁₀ 5; drop the logarithms

1 − log₁₀ 2 = log₁₀ 10 − log₁₀ 2 = log₁₀ 5.
So 5^x + 4^x + 3^x + 2^x + 1 = 5.
At x = 0 each power is 1, so the left side is 5 (holds) (it is increasing in x, so this is the only solution).
x = 0.

Answer: (c) π4

Method. the two terms are reciprocals: t + 1/⁠t = 2 forces t = 1

Let t = log_{sin x} cos x; then log_{cos x} sin x = 1t, and t + 1t = 2 gives (t − 1)² = 0, t = 1.
So cos x = sin x, i.e. tan x = 1.
The bases and arguments must be positive and ≠ 1, so sin x > 0, cos x > 0: x = π4 + 2kπ.
Smallest positive x = π4. (At x = π2 the base sin x = 1 and the logs are undefined.)

No correct option. This question is defective as printed in UPSC's paper.

Method. Vieta's relations with t = log_0.5(a²)

Put t = log_0.5(a²) > 0; then log_a²(0.5) = 1t, and the equation is x² + tx + t⁴ = 0.
Vieta: α + β = -t, αβ = t⁴; the condition is β² = αt, i.e. α = tβ².
Then αβ = tβ³ = t⁴ gives β³ = t³. The real choice β = t, α = t³ makes α + β = t + t³ > 0 ≠ -t, so it fails.
The other cube roots give β = tω or tω²; with β = tω, α = tβ² = t³ω², and α + β = -t becomes t²ω² + ω = -1 = ω + ω², so t² = 1 and t = 1.
So the only consistent case is t = 1 (a² = 0.5): x² + x + 1 = 0, whose roots are the non-real cube roots of unity ω, ω², with β² = α.
So β = ω (or ω²), while options (a) and (b) both equal 1 and (c) and (d) both equal 2 at t = 1: no option is correct.

The data force t = log_0.5(a²) = 1 and non-real roots ω, ω², so β matches no option (and (a), (b) coincide, as do (c), (d)); the evident intended working, β³ = t³ so β = t = log_0.5(a²), gives option (b).

No correct option. This question is defective as printed in UPSC's paper.

Method. Vieta's relations with t = log_0.5(a²)

With t = log_0.5(a²) > 0: α + β = -t, αβ = t⁴, and β² = αt.
As in the previous item, these force t = 1 and {α, β} = {ω, ω²} with α = β².
Then α = 2β would need ω² = 2ω, i.e. ω = 2; α = -2β would need ω = -2; 2α = ±β would need ω = ±12. None holds, since |ω| = 1 and ω is not real.
No option is correct.

The stem's conditions are consistent only for t = 1 with roots ω, ω² (α = β²), which satisfy none of the four relations; the evident intended working, β = t and then α = -t − β = -2β, gives option (c).

Answer: (b) One

Method. bring both logs to base 2, solve, then apply the domain x > 3

Domain: x − 1 > 0 and x − 3 > 0, so x > 3.
log₄(x − 1) = 12 log₂(x − 1), so the equation is log₂(x − 1) = 2 log₂(x − 3), i.e. x − 1 = (x − 3)².
x² - 7x + 10 = 0, so (x − 2)(x − 5) = 0 and x = 2 or 5.
x = 2 fails x > 3; x = 5 works: log₄ 4 = 1 = log₂ 2.
Exactly one solution.

Answer: (c) π4

Method. log_b a = 1 means a = b

log_{cos x} sin x = 1 means sin x = (cos x)¹ = cos x.
On 0 < x < π2, tan x = 1 gives x = π4.
Check: the base cos π4 = 1⁠√2⁠ is positive and ≠ 1, and the argument sin π4 > 0.

Answer: (c) x ∈ (0, 17) ∪ (7, ∞)

Method. take log₇ of both sides to get a quadratic inequality in log₇ x

Both sides are positive, and log₇ is increasing, so x^{log₇ x} > 7 ⇔ (log₇ x)(log₇ x) > 1.
Put t = log₇ x: t² > 1, so t < -1 or t > 1.
t < -1 gives 0 < x < 17; t > 1 gives x > 7.
x ∈ (0, 17) ∪ (7, ∞).

Shortcut. Test x = 149: (149)^-2 = 2401 > 7, which rules out (b) and (d); x = 1 gives 1 > 7, false, which rules out (a).

Answer: (c) 1

Method. write x as log₁₅ 15^x and drop the logarithms

x = x log₁₅ 15, so log₁₅[15^x(1 + 3^x)] = log₁₅(5^x · 12).
Hence 15^x(1 + 3^x) = 12 · 5^x; divide by 5^x: 3^x(1 + 3^x) = 12.
Put t = 3^x > 0: t² + t − 12 = 0, so (t + 4)(t − 3) = 0 and t = 3.
3^x = 3 gives x = 1.

Answer: (c) 1

Method. write x as log₁₀ 10^x and drop the logarithms

x = x log₁₀ 10, so the equation is log₁₀[10^x(1 + 2^x)] = log₁₀(5^x · 6).
Hence 10^x(1 + 2^x) = 6 · 5^x; divide by 5^x: 2^x(1 + 2^x) = 6.
Put t = 2^x > 0: t² + t − 6 = 0, so (t + 3)(t − 2) = 0 and t = 2 (t = -3 is impossible).
2^x = 2 gives x = 1.

Shortcut. Try the options: x = 1 gives 1 + log₁₀ 3 = log₁₀ 5 + log₁₀ 6 = log₁₀ 30 (holds).

Answer: (b) II only

Method. take ln of p^x = q^y = r^z = k and use y² = xz

Let p^x = q^y = r^z = k; then x = ln kln p, y = ln kln q, z = ln kln r.
x, y, z in GP: y² = xz, so (ln k)²(ln q)² = (ln k)²ln p · ln r, i.e. (ln q)² = ln p · ln r.
So ln p, ln q, ln r are in GP: II is correct.
I fails in general: ln p = 1, ln q = 2, ln r = 4 (x = 8, y = 4, z = 2 with k = e⁸) gives p = e, q = e², r = e⁴, which are not in AP.
Only II is correct.

Answer: (b) II only

Method. AP of logs gives y² = xz

Subtracting 10 from each term, log₁₀ x, log₁₀ y, log₁₀ z are in AP.
So 2 log₁₀ y = log₁₀ x + log₁₀ z: the AM of log₁₀ x and log₁₀ z is log₁₀ y, so II is correct.
Also y² = xz, so the GM of x and z is ⁠√(xz)⁠ = y, not y² (e.g. x = 2, y = 6, z = 18: GM = 6, y² = 36). I is false.
Only II is correct.

Answer: (c) 9 < n < 12

Method. log₁₀ 2^j + log₁₀ 5^j = j, so p + q = 1 + 2 + … + n

p + q = nΣj=1 (log₁₀ 2^j + log₁₀ 5^j) = nΣj=1 log₁₀ 10^j = nΣj=1 j = n(n + 1)2.
n(n + 1)2 = 66 gives n(n + 1) = 132 = 11 × 12, so n = 11.
9 < 11 < 12.

Answer: (d) 15log₁₀ 2.5

Method. p + q = n(n + 1)/⁠2 fixes n; q − p = (n(n + 1)/⁠2) log₁₀ 2.5

As in the previous item, p + q = n(n + 1)2 = 15, so n(n + 1) = 30 and n = 5.
q − p = nΣj=1 j(log₁₀ 5 − log₁₀ 2) = n(n + 1)2 log₁₀ 52.
= 15 log₁₀ 2.5.

Answer: (c) Both I and II

Method. ln(xⁿ) = n ln x; test the AP and GP conditions

p = ln x, q = 3 ln x, r = 5 ln x, with ln x > 0 since x > 1.
q − p = r − q = 2 ln x, so p, q, r are in AP: I is correct.
For a GP, q² = pr: 9(ln x)² = 5(ln x)², which needs ln x = 0, i.e. x = 1, excluded.
So p, q, r are never in GP: II is also correct.
Both I and II are correct.

Answer: (c) log_a(log_b a)2

Method. square of the middle term of a GP equals the product of the outer terms; log_x a · log_b x = log_b a

(a^x)² = log_x a · log_b x.
By change of base, log_x a · log_b x = log alog x · log xlog b = log_b a.
So a^2x = log_b a; take log_a: 2x = log_a(log_b a).
x = log_a(log_b a)2.

Answer: (c) log₂ 5

Method. middle term of an AP is the average; drop the logs and solve a quadratic in 2^x

2 log₁₀(2^x - 1) = log₁₀ 2 + log₁₀(2^x + 3), so (2^x - 1)² = 2(2^x + 3).
Put t = 2^x: t² - 2t + 1 = 2t + 6, so t² - 4t − 5 = 0, (t − 5)(t + 1) = 0.
t = -1 is impossible (2^x > 0), and the log needs 2^x - 1 > 0, which t = 5 satisfies.
2^x = 5 gives x = log₂ 5.

Shortcut. Check: the terms are log₁₀ 2, log₁₀ 4, log₁₀ 8, with common difference log₁₀ 2.