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Logarithm: Practice Problems
Solve the given practice questions based on Logarithm. Also, the answer key and explanations are given for the same.
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Q.1.
Find the value of log
9
59049
A. 9
B. 7
C. 5
D. 8
Answer & Explanation
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Sol : Option
B
We have log
9
59049 = log
9
9
5
= 5 × log
9
9 = 5 × 1 = 5.
Q.2.
. If log 2 = 03.301 and log 3 = 0.4771, find the value of log3 72
5
A. 19.46
B. 18.96
C. 21.54
D. 14.48
Answer & Explanation
Sol : Option
A
Q.3.
If x, y and z are the sides of a right angled triangle, where ‘z’ is the hypotenuse, then find the value of (1/log
x+z
y) + (1/log
x-z
y)
A. 1
B. 2
C. 3
D. 4
Answer & Explanation
Sol : Option
B
Here x, y and z are the sides of a right angled triangle, so z
2
= x
2
+ y
2
.
Q.4.
Find the value of log
2
2 + log
2
2
2
+ log
2
2
3
+ ........ + log
2
2
n
.
A. n(n+1)/2
B. n+1
C. n
D. 2n
Answer & Explanation
Sol : Option
A
log
2
2 + log
2
2
2
+ log
2
2
3
+ ........ + log
2
2
n
log
2
2 + 2log
2
2 + 3log
2
2 + ........ + nlog
2
2
1+2+3+........+n
n(n+1)/2
Q.5.
If log
5
16, log
5
(3
x
-4), log
5
(3
x
+97/16) are in arithmetic progression, then x is
A. 8
B. 1
C. 5
D. 3
Answer & Explanation
Sol : Option
B
log
5
16, log
5
(3
x
-4), log
5
(3
x
+97/16) are in arithmetic progression
Must Read Logarithms Articles
Logarithms: Concepts & Theory
Logarithm: Solved Examples
Logarithm: Practice Problems
Q.6.
If ‘x’ is an integer then solve (log
2
x)
2
– log
2
x
4
- 32 = 0.
A. 125
B. 256
C. 375
D. None of these
Answer & Explanation
Sol : Option
B
We have (log
2
x)
2
– log
2
x
4
- 32 = 0.
⇒ (log
2
x)
2
– 4log
2
x - 32 = 0......(1)
Let log
2
x = y
(i) ⇒ y
2
– 4y – 32 = 0
⇒ y
2
– 8y + 4y – 32 = 0
⇒ y (y – 8) + 4 (y – 8) = 0
⇒ (y – 8) (y + 4) = 0
⇒ y = 8, -4
⇒ log
2
x = 8 or log
2
x = - 4
⇒ x = 2
8
= 256 or x = 2
-4
= 1/16
Since ‘x’ is an integer so x = 256.
Q.7.
If log
5
y – logsub>5√y = 2 log
y
5, then find the value of y.
A. 25
B. 35
C. 10
D. 15
Answer & Explanation
Sol : Option
A
We have log
5
y – logsub>5√y = 2 log
y
5
Q8.
If (1/4)log
2
x + 4log
2
y = 2 + log
64-1
8 then
A. y
16
= 64/x
2
B. x
16
= 64/y
C. y
16
= 8/x
4
D. y
16
= 64/x
Answer & Explanation
Sol : Option
D
We have (1/4)log
2
x + 4log
2
y = 2 + log
64-1
8
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Q9.
If log 2 = 0.301 and log 3 = 0.4771, find the number of digits in 48
12
.
A. 19
B. 21
C. 20
D. 24
Answer & Explanation
Sol : Option
B
We have log 48
12
= 12 × log 48 = 12 × log (2
4
× 3)
= 12 × (4 log 2 + log 3)
= 12 × (4 × 0.301 + 0.4771)
= 12 × (1.204 + 0.4771)
= 12 × 1.6811 = 20.1732
Now the characteristic is 20, so the number of digits = 20 + 1 = 21.
Q10.
If log
3
[log
2
(x
2
– 4x – 37)] = 1, where ‘x’ is a natural number, find the value of x.
A. 9
B. 7
C. 10
D. 4
Answer & Explanation
Sol : Option
A
We have log
3
[log
2
(x
2
– 4x – 37)] = 1
⇒ [log
2
(x
2
– 4x – 37)] = 3
⇒ x
2
– 4x – 37 = 8
⇒ x
2
– 4x – 45 = 0
⇒ (x – 9) (x + 5) = 0
⇒ x = 9, - 5
Since x is a natural number, so x = 9.
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