Algebra - Logarithms - Previous Year CAT/MBA Questions
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The number of distinct integer values of n satisfying < 0, is
Answer: 47
Explanation :
Given, < 0
Case 1: 4 – log2n < 0 and 3 – log4n > 0
⇒ log2n > 4 and log4n < 3
⇒ n > 16 and n < 64
∴ integral values of n can be 17, 18, …, 63 i.e., 47 values.
Case 2: 4 – log2n > 0 and 3 – log4n < 0
⇒ log2n < 4 and log4n > 3
⇒ n < 16 and n > 64
∴ No integral values of n is possible.
Hence, 47.
Workspace:
If 5 – + 4 = , then 100x equals
Answer: 99
Explanation :
Given,
5 – + 4 =
⇒ 5 – + =
⇒ 5 = - -
⇒ 5 =
⇒ 5 =
⇒ 5 = -5/2 × log10(1 - x)
⇒ -2 = log10(1 - x)
⇒ 1 – x = 10-2 = 1/100
⇒ 100 – 100x = 1
⇒ 100x = 99
Hence, 99.
Workspace:
log2 [3 + log3 {4 + log4 (x - 1)}] - 2 = 0, then 4x equals
Answer: 5
Explanation :
Given, log2[3 + log3{4 + log4(x - 1)}] - 2 = 0
⇒ log2[3 + log3{4 + log4(x - 1)}] = 2
⇒ 3 + log3{4 + log4(x - 1)} = 4
⇒ log3{4 + log4(x - 1)} = 1
⇒ 4 + log4(x - 1) = 3
⇒ log4(x - 1) = -1
⇒ x – 1 = ¼
⇒ x = 5/4
⇒ 4x = 5
Hence, 5.
Workspace:
For a real number a, if = 4, then a must lie in the range
- A.
4 < a < 5
- B.
2 < a < 3
- C.
a > 5
- D.
3 < a < 4
Answer: Option A
Explanation :
= 4
Converting all logs to base 10.
= 4
⇒ loga32 + loga15 = 4
⇒ loga(32 × 15) = 4
⇒ a4 = 480
This is possible when 4 < a < 5.
Hence, option (a).
Workspace:
If y is a negative number such that , then y equals
- A.
log₂(1/5)
- B.
–log₂ (1/3)
- C.
–log₂ (1/5)
- D.
log₂ (1/3)
Answer: Option D
Explanation :
Given, ,
Taking log on both sides (Choosing the base to be 3)
⇒ y2 × log35 × log32 = log23 × log35
⇒ y2 × log32 = log23
⇒ y2 = (log23)2
⇒ y = - log23 (∵ y is a negative number)
⇒ y = log2(1/3)
Hence, option (d).
Workspace:
If log₄ 5 = (log₄ y)(log₆ √5), then y equals
Answer: 36
Explanation :
Given, log45 = log4y × log6√5
⇒ log45 × logy4 = log6√5
⇒ logy5 = ½ × log65
⇒ logy5 = log365
⇒ y = 36.
Hence, 36.
Workspace:
The value of , for 1 < a ≤ b cannot be equal to
- A.
0
- B.
-1
- C.
1
- D.
-0.5
Answer: Option C
Explanation :
= 1 - logab + 1 - logba = 2 – (logab + logba)
We know logab is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e.,
∴ logab + logba ≥ 2
⇒ 2 – (logab + logba) ≤ 2 – 2
⇒ 2 – (logab + logba) ≤ 0
∴ 2 – (logab + logba) cannot take positive value i.e., it cannot be equal to 1 (option (b))
Hence, option (c).
Workspace:
equals
Answer: 24
Explanation :
Given,
=
=
=
= 23 × 3 = 24
Hence, 24.
Workspace:
If loga30 = A, loga(5/3) = -B and log2a = 1/3, then log3a equals
- A.
(A + B)/2 - 3
- B.
2/(A + B) - 3
- C.
2/(A + B - 3)
- D.
(A + B - 3)/2
Answer: Option C
Explanation :
Since all the 4 options have A + B, lets add A and B.
∴ A + B = loga30 + (-loga(5/3))
= loga30 – loga(5/3) = loga(30 × 3/5)
= loga(18) = loga(2 × 32)
⇒ A + B = loga(2) + 2loga(3)
⇒ A + B = 3 + 2loga(3) [∵ log2a = 1/3 ∴ loga2 = 3]
⇒ loga(3) = (A + B - 3)/2
⇒ log3(a) = 2/(A + B - 3)
Hence, option (c).
Workspace:
Let x and y be positive real numbers such that log5(x + y) + log5(x - y) = 3, and log2y - log2x = 1 - log23. Then xy equals
- A.
250
- B.
150
- C.
100
- D.
25
Answer: Option B
Explanation :
log5(x + y) + log5(x - y) = 3
∴ log5(x + y)(x - y) = 3.
∴ (x + y)(x − y) = 53 = 125
So, x2 - y2 = 125 ....(1)
log2y - log2x = 1 - log23
∴ = 1
∴ (3y/x) = 21 = 2.
So, 3y = 2x .... (2)
Solving (1) and (2), we get;
x = 15 and y =10.
∴ xy = 15 × 10 = 150.
Hence, option (b).
Workspace:
The real root of the equation 26x + 23x+2 - 21 = 0 is
- A.
- B.
log29
- C.
- D.
log227
Answer: Option A
Explanation :
Given: 26x + 23x+2 - 21 = 0
Replace 23x with y
So, 26x = y2
Now, 26x + 23x+2 - 21 = 0 can be rewritten as y2 + 22 × 23x - 21 = 0
y2 + 4y - 21 = 0
Solving the above quadratic equation,
(y + 7) (y - 3) = 0
So, y = -7 or +3
y = - 7 is rejected (since, y = 23x which should always be positive)
⇒ 23x = 3
Taking log on both sides,
log23 = 3x
x =
Hence, option (a).
Workspace:
Let A be a real number. Then the roots of the equation x2 - 4x - log2A = 0 are real and distinct if and only if
- A.
A < 1/16
- B.
A < 1/8
- C.
A > 1/8
- D.
A > 1/16
Answer: Option D
Explanation :
For roots of a quadratic equation to be real and distinct, Discriminant > 0
So, for x2 − 4x – log2A = 0,
D = (-4)2 - (4 × 1 × (-log2A)) > 0
⇒ 16 + 4 × log2A > 0
⇒ log2A > -4
⇒ A > 2-4
⇒ A > 1/16
Hence, option (c).
Workspace:
If x is a real number, then is a real number if and only if
- A.
-3 ≤ x ≤ 3
- B.
1 ≤ x ≤ 2
- C.
1 ≤ x ≤ 3
- D.
-1 ≤ x ≤ 3
Answer: Option C
Explanation :
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]
Hence, option (c).
Workspace:
If x is a positive quantity such that 2x = , then x is equal to
- A.
1 +
- B.
log59
- C.
log58
- D.
1 +
Answer: Option D
Explanation :
Given, 2x =
⇒ 2x =
⇒ x = log53
Option (a): 1 + = 1 + log35 - log33 = log35
Option (b): log59 = 2(log53)
Option (c): log58 = 3(log52)
Option (d): 1 + = 1 + log53 - log55 = log53
Hence, option (d).
Workspace:
If log1281 = p, then is equal to
- A.
log28
- B.
log616
- C.
log68
- D.
log416
Answer: Option C
Explanation :
log1281 = p
⇒ p = 4log123
Now,
=
=
= =
Hence,
Alternately,
log1281 = p
⇒ 81 = 12p
⇒ 34 = 3p × 22p
⇒ 3(4 − p) = 22p
Taking log on both the sides,
(4 – p) (log 3) = (2p) (log 2)
∴ =
∴ = …(by adding 1 both sides)
∴ =
∴ = = log62
∴ = 3log62 = log68
Hence, option (c).
Workspace:
If log2(5 + log3 a) = 3 and log5(4a + 12 + log2 b) = 3, then a + b is equal to
- A.
40
- B.
67
- C.
59
- D.
32
Answer: Option C
Explanation :
log2(5 + log3a) = 3
⇒ (5 + log3a) = 23 = 8
⇒ log3a = 3
⇒ a = 33 = 27
log5(4a + 12 + log2b) = 3
⇒ log5(4 × 27 + 12 + log2b) = 3
⇒ 120 + log2b = 53
⇒ log2b = 125 - 120
⇒ log2b = 5
⇒ b = 25 = 32
∴ a + b = 27 + 32 = 59
Hence, option (c).
Workspace:
If p3 = q4 = r5 = s6, then the value of logs(pqr) is equal to
- A.
47/10
- B.
16/5
- C.
24/5
- D.
1
Answer: Option A
Explanation :
Given, p3 = q4 = r5 = s6 = k
⇒ p3 = k
⇒ p = k1/3
Similarly,
q = k1/4
r = k1/5
s = k1/6
pqr = k1/3 × k1/4 × k1/5 = k1/3 + 1/4 + 1/5 = k47/60
Now, logs(pqr) = logk1/6 k47/60 = 47/10
Hence, option (a).
Workspace:
The smallest integer n for which 4n > 1719 holds, is closest to
- A.
33
- B.
37
- C.
39
- D.
35
Answer: Option C
Explanation :
We have, 4n > 1719
Taking log on both sides to the base 4,
n × log4 4 > 19 × log4 17
Now, 17 > 42.
∴ log4 17 > 2.
⇒ n × log4 4 > 19 × log4 17 > 19 × 2
Also, log4 4 = 1
∴ n > 38.
Hence, option (c).
Workspace:
- A.
-4
- B.
10
- C.
0
- D.
1/2
Answer: Option D
Explanation :
- + - + - +
= - + - + - +
= = =
Hence, option (d).
Workspace:
Suppose, log3x = log12y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log6G is equal to:
- A.
√a
- B.
2a
- C.
a/2
- D.
a
Answer: Option D
Explanation :
log3x = a ⇒ x = 3a.
log12y = a ⇒ y = 12a.
∴ xy = 36a and √(xy) = G = 6a.
∴ log6G = a.
Hence, option (d).
Workspace:
The value of log0.008√5 + log√381 – 7 is equal to:
- A.
13
- B.
23
- C.
56
- D.
76
Answer: Option C
Explanation :
0.008 = 8/1000 = 1/125 = 5-3
∴ log0.008√5 = = = and
log√381 = = = 8.
∴ The given expression is -1/6 + 8 - 7 = 5/6
Hence, option (c).
Workspace:
If x is a real number such that log35 = log5(2 + x), then which of the following is true?
- A.
0 < x < 3
- B.
23 < x < 30
- C.
x > 30
- D.
3 < x < 23
Answer: Option D
Explanation :
Given log35 = log5(x + 2)
Now log33 < log35 < log39
∴ 1 < log35 < 2
⇒ 1 < log5(x + 2) < 2
⇒ log55 < log5(x + 2) < 2log55
⇒ log55 < log5(x + 2) < log525
⇒ 5 < 2 + x < 25
⇒ 3 < x < 23
Hence, option (d).
Workspace:
If log (2a × 3b × 5c) is the arithmetic mean of log (22 × 33 × 5), log (26 × 3 × 57), and log (2 × 32 × 54), then a equals
Answer: 3
Explanation :
Arithmetic mean of log(22 × 33 × 5), (log 26 × 3 × 57) and (log 2 × 32 × 54)
= [log(22 × 33 × 5) + log(26 × 3 × 57) + log(log 2 × 32 × 54)]
= [log(22 × 33 × 5 × 26 × 3 × 57 × 2 × 32 × 54)]
= [log(22 + 6 + 1) × (33 + 1 + 2) × (51 + 7 + 4)]
=
= log [23 × 32 × 54]
Now log(2a × 3b × 5c) = log(23 × 32 × 54)
(Note: In the given question it should have been specified that a, b, c are integers, as there can be multiple answers for the same. Since it is a TITA question, we have to assume the same and equate power of 2, 3 and 5)
∴ Equating powers of 2, 3 and 5 we get
a = 3, b = 2 and c = 4
Hence, 3.
Workspace:
If logyx = a × logzy = b × logxz = ab, then which of the following pairs of values for (a, b) is not possible?
- A.
- B.
- C.
0.4, 2.5
- D.
- E.
2, 2
Answer: Option E
Explanation :
If logyx = a × logzy = b × logxz = ab
⇒ logyx = ab ...(1)
⇒ a × logzy = ab ⇒ logzy = b ...(2)
⇒ b × logxz = ab ⇒ logxz = a ...(3)
From (1), (2) and (3), we get
logyx = logxz × logzy
∴ = =
∴ (log x)2 = (log y)2
∴ log x = ± log y
∴ log x = log y or log x = - log y
∴ x = y or x =
∴ ab = logy x = 1 or -1
Only option (e) does not satisfy this.
Hence, option (e).
Workspace:
If x ≥ y and y > 1, then the value of the expression
can never be
- A.
-1
- B.
-0.5
- C.
0
- D.
1
Answer: Option D
Explanation :
= logx x - logx y + logy y - logy x
⇒ = 1 - logx y + 1 - logy x
⇒ = 2 - logx y - logy x
⇒ = 2 - (logx y + logy x)
As x ≥ y and y > 1,
logy x ≥ 0
Now, (logx y + logy x) = [This is sum of a positive number and its reciprocal]
Now, sum of a positive number and its reciprocal is always greater than or equal to 2.
∴ (logx y + logy x) = ≥ 2
⇒ = 2 - (logx y + logy x) ≤ 0
∴ ≠ 1
Hence, option (d).
Workspace:
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