Algebra - Logarithms - Previous Year CAT/MBA Questions
The best way to prepare for Algebra - Logarithms is by going through the previous year Algebra - Logarithms cat questions. Here we bring you all previous year Algebra - Logarithms cat questions along with detailed solutions.
Click here for previous year questions of other topics.
It would be best if you clear your concepts before you practice previous year Algebra - Logarithms cat questions.
If log4 m + log4 n = log2 (m + n) where m and n are positive real numbers, then which of the following must be true?
- (a)
+ = 1
- (b)
m = n
- (c)
m2 + n2 = 1
- (d)
+ = 2
- (e)
No values of m and n can satisfy the given equation
Answer: Option E
Explanation :
log4 mn = log2 (m + n)
= (m + n)
Squaring on both sides
m2 + n2 + mn = 0
Since m, n are positive real numbers, no value of m and n satisfy the above equations.
Workspace:
The value of x which satisfy 6 - 9 log8 - 8( - = 0 is
- (a)
2
- (b)
- (c)
4
- (d)
8
Answer: Option D
Explanation :
6 - 9 log8 - 8 - = 0
6 - log2 4 + log2 x - 2 - 2 = 0
Let be t.
4 + t3 - 2t2 - 2t = 0
or, (t - 2) (t2 - 2) = 0
so, t = 2 or
log2 x = 8
Workspace:
If log3 = 243, then what will be the value of x:
- (a)
x ∈ Q rational numbers
- (b)
x ∈ N Natural numbers
` - (c)
x ∈ N Even natural numbers
- (d)
x ∈ Q rational numbers, x < 0
Answer: Option B
Explanation :
Workspace:
Find the value of following expression log sin 40° log sin 41° ---- log sin 90° log sin 100°
- (a)
- (b)
0
- (c)
1
- (d)
2
Answer: Option B
Explanation :
= log sin 40° log sin 41° ---- log sin 90° log sin 100°
= log sin 90°
= log 1
= 0
Answer is option B.
Workspace:
equals which of the following?
- (a)
-5
- (b)
-3
- (c)
None of the others
- (d)
-4
- (e)
-2
Answer: Option D
Explanation :
Given
=
Now, =
⇒ = = = -4
Workspace:
- (a)
0
- (b)
1
- (c)
xyz
- (d)
Answer: Option B
Explanation :
Since there is no condition imposed on x, y and z, assume x = y = z = 10
Consider logxyz + 1.
When x = y = z = 10; logxyz + 1
= log10(10 × 10) + 1
= log10(10)2 + 1 = 2 log10(10) + 1
= 2(1) + 1 = 3
Similarly, logyxz + 1 = logzxy + 1 = 3
∴ Required value = (1/3) + (1/3) + (1/3) = 1
Hence, option (b).
Workspace:
- (a)
12
- (b)
244
- (c)
243
- (d)
245
Answer: Option B
Explanation :
Consider loge3 = x.
Hence, the series can also be written as
Now, each bracket is a standard expression that can be expressed as a power of e. Hence, the expression becomes:
= 3 – 1 + 35 – 1 = 1 + 243 = 244
Hence, option (b).
Note:
This is one of the rare questions where pure engineering maths based concepts have been used. In the exam, you are advised to leave such a question if you are not conversant with such series.
Workspace:
Find the value of x which satisfies the following equation
4 log7 (x - 8) = log3 81
- (a)
8
- (b)
18
- (c)
20
- (d)
None of the above
Answer: Option D
Explanation :
4 log7(x – 8) = log3 81
∴ 4 log7(x – 8) = log3(34) = 4
∴ log7(x – 8) = 1
∴ x – 8 = 7 i.e. x = 15
Hence, option (d).
Workspace:
If log255 = a and log2515 = b, then the value of log2527 is:
- (a)
3(b + a)
- (b)
3(1 - b - a)
- (c)
3(a + b - 1)
- (d)
3(1 - b + a)
Answer: Option C
Explanation :
a + b = log255 + log2515 = log2575 = log2525 + log253
a + b = 1 + log253
a + b – 1 = log253
∴ log2527 = 3log253 = 3(a + b – 1)
Hence, option (c).
Workspace:
If x - = 6 10 then the value of x is
- (a)
10
- (b)
30
- (c)
100
- (d)
1000
Answer: Option D
Explanation :
x - =
∴ x - x =
x = a
∴ a2 - a2 = 6
∴ a2 = 9
∴ a = ±3
As log cannot be negative a = 3
∴ x = 3
∴ x = 1000
Hence, option (d).
Workspace:
If log13 log21 { + } = 0 then the value of x is
- (a)
21
- (b)
13
- (c)
81
- (d)
None of the above
Answer: Option D
Explanation :
log13 log21 { + } = 0
∴ log21 { + } = 13° = 1
∴ { + } = 211 = 21
∴ = 21 -
Squaring both sides,
x + 21 = 441 + x - 42
∴ 42 = 420
∴ x = 100
Hence, option (d).
Workspace:
If log 3, log (3x – 2) and log (3x + 4) are in arithmetic progression, then x is equal to
- (a)
- (b)
log38
- (c)
log23
- (d)
8
Answer: Option B
Explanation :
log(3x – 2) – log3 = log (3x + 4) – log(3x – 2)
Let 3x = t
∴ log = log
∴ =
∴t2 + 4 – 4t = 3t + 12
∴ t2 – 7t – 8 = 0
∴ (3x – 8)(3x + 1) = 0
Since 3x cannot be negative,
∴ 3x = 8
∴ x = log38
Hence, option (b).
Workspace:
Find the value of x from the following equation:
Log103 + log10 (4x+1) = log10 (x+1) + 1
- (a)
2/7
- (b)
7/2
- (c)
9/2
- (d)
None of the above
Answer: Option B
Explanation :
log10 3 + log10 (4x + 1) = log10 (x + 1) + 1
∴ log10 3 + log10 (4x + 1) – log10 (x + 1) = 1
∴ log10 [3(4x + 1)/(x + 1)] = 1
∴ (12x + 3)/(x + 1) = 10
∴ x = 7/2
Hence, option (b).
Workspace:
What is the value of ,
If log4 log4 4a - b = 2 log4 ( - ) + 1
- (a)
-5/3
- (b)
2
- (c)
5/3
- (d)
1
Answer: Option C
Explanation :
log4 log4 4a - b = 2log4 + 1
∴ log4 (a - b) log4 4 = 2log4 + 1
∴ log4 (a - b) = 2log4 + 1
∴ log4 (a - b) = 2log4 + 1
∴ log4 = 1
∴ = 4
∴ a - b = 4a + 4b - 8
∴ 8 = 3a + 5b
∴ 8 = 3 + 5
Let = x
∴ 8 = 3x +
∴ 8x = 3x2 + 5
∴ 3x2 - 8x + 5 = 0
∴ 3x2 – 3x – 5x + 5 = 0
∴ 3x(x – 1) – 5(x – 1) = 0
∴(3x – 5)(x – 1) = 0
∴ x = or x = 1
But x ≠ 1 as ≠
∴ x ≠ 5/3
∴ =
Hence, option (c).
Workspace:
log5 2 is
- (a)
An integer
- (b)
A rational number
- (c)
A prime number
- (d)
An irrational number
Answer: Option D
Explanation :
Let log5 2 be rational.
Then log5 2 =
∴ = 2
∴ =
∴ 5m = 2n
However, number 2 raised to any positive integer power must be even, but 5 raised to any positive integer power must be odd.
Hence, we have a contradiction.
∴ log5 2 is irrational.
Hence, option (d).
Workspace:
The sum of the series is:
+ + + ....
- (a)
e2 – 1
- (b)
loge2 – 1
- (c)
2log102 – 1
- (d)
None of the above
Answer: Option D
Explanation :
S =
=
=
=
= - ...(i)
Now, =
=
=
= -
= - + ...(ii)
From (i) and (ii), we get,
S = - +
= -
= - + ...(iii)
Now, we know that,
loge (1 + x) = x - x2 + x3 - x4 + ...
= (-1)n- 1 xn
Putting x = 1 in the above equation, we get,
loge 2 = ...(iv)
From (iii) and (iv), we get,
S = loge 2 -
Hence, option (d).
Alternatively,
You can also solve this question using options.
+ + + ... = + + + ...
= 0.167 + 0.0167 + 0.0047 + …
< 0.2 but always positive
So, value of S will always be less than 0.2.
Now, we will consider all the given options one by one.
Consider option 1.
e2 – 1 = (2.718)2 – 1 > 0.2
So, option 1 can be eliminated.
Consider option 2.
loge2 – 1 = 0.693 – 1 < 0
So, option 2 can be eliminated.
Consider option 3.
2log102 – 1 = 2(0.3010) – 1 < 0
So, option 3 can be eliminated.
So, the correct answer will be option 4.
Hence, option (d).
Workspace:
If log2x.log2 = log2. Then x is:
- (a)
2
- (b)
4
- (c)
16
- (d)
12
Answer: Option B
Explanation :
log2x.log2 = log2
∴ × =
∴ = log 2
∴ (log x) × = log 2
Let log x = t
∴ = log 2
∴ t2 - t log 16 = t log 2 - log 2. log 64
∴ t2 - 4t log 2 = t log 2 - 6(log 2)2
∴ t2 - 5t log 2 + 6(log 2)2 = 0
∴ (t - 2 log 2) (t - 3 log 2) = 0
∴ t = log 4 or t = log 8
∴ x = 4 or x = 8
Hence, option (b).
Workspace:
Feedback
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.