A = log₂₇625 + 7log₁₁¹³
= log_3³5⁴ + 7 log₁₁13
= 4/3 log₃ 5 + 7 log₁₁13
B = log₉125 + 13 log₁₁7 = log₃² 5³ + 13 log₁₁⁷
= 3/2 log₃ 5 +13 log₁₁ 7
Let log₃ 5 = x and by the above rule
7 log₁₁ 13 = 13 log₁₁ 7
Therefore, A = 4/3 x + 13 log₁₁ 7
and B = 3/2 x + 13 log₁₁ 17
clearly, A < B hence (B) is the correct answer.

Correct Option: B

A = log₂₇625 + 7log₁₁¹³
= log_3³5⁴ + 7 log₁₁13
= 4/3 log₃ 5 + 7 log₁₁13
B = log₉125 + 13 log₁₁7 = log₃² 5³ + 13 log₁₁⁷
= 3/2 log₃ 5 +13 log₁₁ 7
Let log₃ 5 = x and by the above rule
7 log₁₁ 13 = 13 log₁₁ 7
Therefore, A = 4/3 x + 13 log₁₁ 7
and B = 3/2 x + 13 log₁₁ 17
clearly, A < B hence (B) is the correct answer.

∵ log₁₂27 = a
⇒ log 27 / log 12 = a
⇒ a log 12 = log 3³
⇒ a log ( 3 x 4 ) = 3 log 3
⇒ a[log 3 + log 4] = 3 log 3
⇒ a log 4 + a log 3 = 3 log 3
⇒ a log 2² = ( 3 - a) log 3
⇒ 2a log 2 = (3 - a) log 3
∴ log 2 / log 3 = (3 - a) / 2a
Now log₆16 = log16 / log 6 = log 2⁴ / log (2 x 3) = 4 log 2 / ( log 2 + log 3)
= [4 (log 2 / log 3)] / [(log 2 / log 3) + 1]
= 4[(3 - a) / 2a] / [{(3 - a) / 2a } + 1]
= 4(3 - a) / (3 + a)

Correct Option: A

∵ log₁₂27 = a
⇒ log 27 / log 12 = a
⇒ a log 12 = log 3³
⇒ a log ( 3 x 4 ) = 3 log 3
⇒ a[log 3 + log 4] = 3 log 3
⇒ a log 4 + a log 3 = 3 log 3
⇒ a log 2² = ( 3 - a) log 3
⇒ 2a log 2 = (3 - a) log 3
∴ log 2 / log 3 = (3 - a) / 2a
Now log₆16 = log16 / log 6 = log 2⁴ / log (2 x 3) = 4 log 2 / ( log 2 + log 3)
= [4 (log 2 / log 3)] / [(log 2 / log 3) + 1]
= 4[(3 - a) / 2a] / [{(3 - a) / 2a } + 1]
= 4(3 - a) / (3 + a)

log_x4 = log 4 / log x = 2/5
⇒ 2log2 / log x = 2/5
⇒ log x =5log 2 = log 2⁵
⇒ log x = log 32

Correct Option: D

log_x4 = log 4 / log x = 2/5
⇒ 2log2 / log x = 2/5
⇒ log x =5log 2 = log 2⁵
⇒ log x = log 32
∴ x = 32

log_xy =100, log₂x = 10
⇒ log y / log x = 100 and log x / log 2 = 10

Correct Option: B

log_xy =100, log₂x = 10
⇒ log y / log x = 100 and log x / log 2 = 10
⇒ log y / log 2 = 100 x 10 = 1000
⇒ log₂y = 1000
∴ y = 2¹⁰⁰⁰

∵ a^x = b
⇒ log_a b = x

∵ b^y = c
⇒ log_b c = y

∵ c^z = a
⇒ log_c a = z

Correct Option: B

∵ a^x = b
⇒ log_a b = x

∵ b^y = c
⇒ log_b c = y

∵ c^z = a
⇒ log_c a = z

∴ xyz = log_ab x log_bc x log_ca = 1