Find the number of digits in 8 ¹⁰? (Given that log ₁₀ 2 = 0.3010)

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.

log_an / log_abn = [( log n / log a) / (log n / log (a.b))]
= log (a.b) / log a
= ( log a + log b) / log a
= 1 + (log b / log a)
= 1 + log_ab

? 10^x = 1730/1000
? log10^x= log₁₀1730 - log₁₀1000
? x = 3.2380 - 3
= 0.2380

Given Exp. = log₉27 - log₂₇9
= log27 / log9 - log9 / log27
= 3log3 / 2log3 - 2log3 / 3log3
= 3/2 - 2/3 = 5/6

(log₅3) x log₃ 5⁴ = [log3 / log5] x 4 x [log5 / log3]
= 4

8⁵⁷ = (2³)⁵⁷ = 2¹⁷¹
? Required answer = (171 log₁₀ 2 + 1 )
= [171 x 0.3010] + 1 = [51.4710] +1
= 51+1 =52

Given log( x - 5) = log(x) - log(5)
? x -5 = x/5
? x = 25/ 4 ....(i)

Again from question
log(y - 6) = log(y) - log(6)
? y - 6 = y/6
? y = 36/5 ...(ii)

From equations (i) & (ii) x < y

Given log (x + 4) = log(4) + log(x)
? x + 4 = 4x
? x = 4/3

Similarly y = 5/4

? x > y

? 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

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