CAT Previous Year Questions: Logarithms - Quantitative Aptitude (Quant)

We have,

We get,

we get log a(log32 +log15) = 4(log a)²we get (log32 + log 15) = 4loga= log480=loga⁴
 = a⁴  = 480
so we can say a is between 4 and 5 .
 

2^y2log₃ 5 = 5^log₂ 3

 ( is negative )

= log_a a - log_ab +log_bb - log_ba
= 
since ( log_ab + log_ab ) ≥ 2
∴ 1 can't be the answer.

Let the total marks be T and scores of Bishnu, Asha and Ramesh be a, b and c respectively.
Given, a = 52% of T = c - 23 and b = 64% of T = c + 34
Hence, (64 - 52)% of T = (c + 34) - (c - 23) = 57
i.e. 12% of T = 57
Hence, score of Geeta = 84% of T = 7  x 57 = 399

Given that, log₅(x + y) + log₅(x - y) = 3
We know log A + log B = log (A x B)
log₅(x + y) + log₅(x - y) = log₅(x² - y²)
log₅(x² - y²) = 3
x² - y² = 5³
x² - y² = 125
Similarly, log₂y - log₂x = 1 - log₂3
log₂ (y/x) = log₂2 - log₂3
log₂(y/x) = log₂(2/3)
3y = 2x
(3/2)y² - y² = 125
(9/4)y² - y² = 125
5/4 y² = 125
y² = 100
y = 10
x = 15
So, xy = 15 x 10 = 150

It is given that, is a real number
Therefore, ≥ 0
≥ 1
4x - x² ≥ 3
x² - 4x + 3 ≤ 0
(x−1) (x-3) ≤ 0
x ∈ [1,3]

Given equation - 2^6x + 2^3x+2 - 21 = 0
The idea is to convert the above equation to a quadratic one.
So, 2^6x = (2^3x)²
Let y = 2^3x
Now, 2^6x + 2^3x+2 - 21 = 0 can be rewritten as + 2(3x)²+ 2².2^3x - 21 = 0
y² + 4y - 21 = 0
Solving the above quadratic equation,
(y + 7) (y - 3) = 0
So, y = -7 or +3
y = 2^3x ⇒ y should always be positive
Therefore, y = -7 is not a valid solution. So, only y = +3 exists.
2^3x = 3
Taking log on both sides,
log₂ 2^3x = log₂ 3
3x = log₂ 3
x = 1/3 log₂ 3, which is the real solution to the given equation

Given, 2^x = 3^log52
Taking log on both sides, we get
log (2^x) = log (3^log52)
x × log(2) = log₅(2) × log(3)

Let us consider the base as 3,

Going through the options, Simplifying (B)
1 + log₅ (3/5)
1 + log₅(3) - log₅(5)
1 + log₅(3) – 1
log₅(3)

Given, log₁₂81 = p
log₁₂(3)⁴ = p
4 (log₁₂(3)) = p
log₁₂(3) = p/4 ...(1)
We need to find the value of 3 × 
This can be rewritten as 3 × 
Replacing (2) with (1), We get




log₃₆64
log₆8
Hence, the answer is log₆8

Given that p³ = q⁴ = r⁵ = s⁶
We have to find the value of log_s (pqr)
Since more variables are given, to avoid confusion assume a new variable to simplify it.
Let us assume this p³ = q⁴ = r⁵ = s⁶ is equal to k^x
so that we can get every values in k
We can rewrite this logs (pqr) as 
Now let us take the LCM of 3 , 4 , 5 and 6 which is equal to 60
Hence p³ = q⁴ = r⁵ = s⁶ = k⁶⁰
Or p = k²⁰ q = k¹⁵ r = k¹² s = k¹⁰

The question is "If p³ = q⁴ = r⁵ = s⁶, then the value of logs (pqr) is equal to" 
Hence, the answer is 47/10

We know that 
⇒log₁₀₀ 2 - log₁₀₀ 4 + log₁₀₀ 5 - log₁₀₀ 10 + log₁₀₀ 20 - log₁₀₀ 25 + log₁₀₀ 50
We also know that (- log_a b) = log_a (1/b)

log₃ x = log₁₂ y = a, where x, y are positive numbers.
log₃ x = a; x = 3^a
log₁₂ y = a; y = 12^a
If G is the geometric mean of x and y, log₆G is equal to has to be found
We know that geometric mean is √xy
√xy = √(36^a)
= √(6^2a) = (6^2a)^1/2
= 6^a
log₆ G = a

We have to find the value of x
⇒  9^{2x – 1} – 81^x-1 = 1944
⇒ 9^{2x – 1} – (9²)^x-1 = 1944
⇒ 9^{2x – 1} – 9^{2x - 2} = 1944
⇒ (9^{2x - 2} × 9) - 9^{2x - 2} = 1944

We can write 243 = 3⁵ = 9^(5/2)
Comparing the powers,
⇒ 2x – 2 = 5/2
⇒ 2x = 9/2
⇒ x = 9/4

log₃ 5 = log₅ (2 + x)
Since log3 3 = 1 and log₃ 9 = 2 , we can say that 2 > log₃ 5 > 1
We need to find the range of x
If log₅ (2 + x) > 1
⇒ 2 + x > 5
⇒ x > 3
If log₅ (2 + x) < 2
⇒ 2 + x < 5²
⇒ 2 + x < 25
⇒ x < 23
Therefore 3 < x < 23

Here we have to find the value of log_0.008 √5 + log_√3 81 – 7.
⟹ log_0.008 √5 + log_√3 81 – 7
log_0.008 √5 can be written in the terms of five and log_√3 81 can be written in the terms of 3.
where √5 = 5^1/2 ⟹ 0.008 = 2³/10³ = 5^-3
⟹ log_0.008 √5 + log_√3 81 – 7

Hence the value of log_0.008√5 + log_√381 – 7 = 5/6

Given that 

⇒ 3^2x × 8 = 2^2x × 27
⇒ 3^{2x - 3} = 2^{2x - 3}
⇒ 2x - 3 = 0
⇒ x = 3/2