Test Level 2: Exponents and Logarithm - 1 - CAT MCQ

3 log_a p/a²)
= 3[loga p – loga a²]  [∵ log m/n = log m – log n]
= 3[log_a p – 2 log_a a]  [∵ log mⁿ = n log m]
= 3[2.3 – 2 × 1]                 [∵ log_m m = 1]
= 3 × 0.3
= 0.9

log (35 - x³)/log (5 - x) = 3
⇒ log (35 - x³) = 3 log (5 - x)
⇒ (35 - x3) = (5 - x)³
i.e. 35 - x³ = 125 - x³ - 15x(5 - x)
⇒ 15x² - 75x + 90 = 0
Or, 3x² - 15x + 18 = 0
⇒ (3x - 9)(x - 2) = 0
⇒ Either x = 3 or x = 2

log x³ = log 162 + log a - log 6a
= log [162a ÷ 6a]
= log 27
log x³ = log 27
i.e. x³ = 27
x = 3

Given: x^3/7 = 423
On taking log of both the sides, the equation becomes:

It is nearest to 18.38.
So, option 3 is correct answer.

log₇ log₅ 
If x = 4, then
log7 log5 
log7 log5 
log7 [log5 (5)] = 0
log₇ (1) = 0 [because loga (a) = 1]
0 = 0 [because log7 (1) = 0]
LHS = RHS
Hence, x = 4

2log(12/24) - log(25/162) + log(4/9)
= 2log(1/2) - log25 + log162 + log4 - log9
= 2log1 - 2log2 - 2log5 + log(3⁴ x 2) + 2log2 - log3²
= -2log2 - 2log5 + 4log3 + log2 + 2log2 - 2log3
= 2log3 - 2log5 + log2
= log9 + log2 - log25
= log(9 × 2) - log25
= log18 - log25

Hence, option (d) is correct.

2log10 x - log_x (0.01)
= 2log₁₀ x - logx 10^-2
= 2log₁₀ x + 2log_x 10
= 2[log₁₀ x + log_x 10] ≥ 2.2 ≥ 4
As A.M. ≥ G.M. of two positive values.

log₁₀(tan 1°) + log₁₀(tan 2°) + log₁0(tan 3°) + … + log₁₀(tan 88°) + log₁₀(tan 89°)
= log₁₀((tan 1°)(tan 89°)) + log₁₀((tan 2°)(tan 88°)) + ... + log10(tan 45°)
= log₁₀((tan 1°)(cot 1°)) + log₁₀((tan 2°)(cot 2°)) + ... + log₁₀(tan 45°)
= log₁₀ 1 + log₁₀ 1 + … + log₁₀ 1 = 0

We use the change of base formula log_b a =  to rewrite f(n) in terms of logarithms to the base 2.
Thus,
f(n) = log2 3 × = log₂ n
As the other logarithm expressions cancel in the product,
Therefore,
= 2 + 3 + 4 + ……. + 10 = 54

By using base change formula:
log₃ 4 = (log₂ 4)/(log₂ 3)
log₄ 5 = (log₂ 5)/(log₂ 4)
...
log₉₉ 100 = (log₂ 100)/(log₂ 99)
Putting these values in the expression, we get the value of the expression as log₂ 100.
Hence, answer option c is correct.