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CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 9 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Logarithm And Antilogarithm
Quiz 9
Multiply
10
4
by
10
2
Report Question
0%
10
8
0%
10
2
0%
10
6
0%
10
−
2
0%
10
3
Evaluate using logarithm table:
28.45
×
3
√
0.3254
32.43
×
5
√
0.3046
Report Question
0%
0.7666
0%
0.7656
0%
0.5686
0%
0.2936
Explanation
Let
y
=
28.45
×
3
√
.3254
32.43
×
3
√
.3046
ln
y
=
ln
28.45
+
ln
3
√
.3254
−
(
ln
32.43
+
ln
5
√
.3046
)
ln
y
=
ln
25.45
+
1
3
ln
.3245
−
ln
32.43
−
1
5
ln
.4046
ln
y
=
3.236
+
(
−
.375
)
−
3.479
−
(
−
.237
)
ln
y
=
−
.381
y
=
anti
ln
(
−
.381
)
y
=
.7656
So, option B is correct.
If
log
2
a
4
=
log
2
b
6
=
log
2
c
3
p
and also
a
3
b
2
c
=
1
, then the value of
p
is equal to
Report Question
0%
−
6
0%
−
7
0%
−
8
0%
−
9
Explanation
log
2
a
4
=
log
2
b
6
=
log
2
c
3
p
=
x
log
2
a
=
4
x
⇒
a
=
2
4
x
log
2
b
=
6
x
⇒
b
=
2
6
x
log
2
c
=
3
p
x
⇒
c
=
2
3
p
x
a
3
b
2
c
=
1
2
12
x
⋅
2
12
x
⋅
2
3
p
x
=
2
0
2
24
x
+
3
p
x
=
2
0
∴
24
x
+
3
p
x
=
0
∴
3
p
x
=
−
24
x
∴
p
=
−
8
Given
l
o
g
3
(
a
)
=
c
and
l
o
g
3
(
b
)
=
2
c
,
a
=
Report Question
0%
3
c
0%
c
+
3
0%
b
2
0%
√
b
0%
b
2
Explanation
Given,
log
3
(
a
)
=
c
and
log
3
(
b
)
=
2
c
∴
a
=
3
c
and
b
=
3
2
c
Now consider
a
2
as follows:
a
2
=
(
3
c
)
2
⇒
a
2
=
3
2
c
⇒
a
2
=
b
⇒
a
=
√
b
The number
N
=
6
log
10
2
+
log
10
31
lies between two successive integers, whose sum is equal to
Report Question
0%
5
0%
7
0%
9
0%
10
Explanation
N
=
6
log
10
2
+
log
10
31
=
log
10
2
6
+
l
o
g
10
(
31
)
=
log
10
(
2
6
.31
)
=
log
10
(
64
×
31
)
=
log
10
(
1984
)
Now
log
10
(
10
3
)
=
log
10
(
1000
)
=
3
Similarly
log
10
(
10
4
)
=
4
Since
10
3
<
1984
<
10
4
3
<
log
10
(
1984
)
<
4
Hence it lies between
3
and
4
.
The sum of these two integers is
7
.
If
log
10
2
=
0.3010
, then the number of digits in
2
64
is
Report Question
0%
18
0%
24
0%
22
0%
20
Explanation
Given
log
10
2
=
0.301
log
10
2
64
=
64
×
log
10
2
=
64
×
0.3010
=
19.264
⇒
2
64
=
10
19.264
The number of digits in
10
19
is
20
, there will be
21
digits from
10
21
The number
10
19.264
lies between them
Therefore the number of digits in
10
19.264
is
20
Therefore the correct option is
D
Approximate of
log
11
21
is
Report Question
0%
1.27
0%
1.21
0%
1.18
0%
1.15
0%
1.02
Explanation
Approximate value of
log
11
21
=
log
11
(
7
×
3
)
=
log
11
7
+
log
11
3
=
0.8115
+
0.4581
=
1.27
2
1
/
4
4
1
/
8
8
1
/
16
16
1
/
32
....... is equal to
Report Question
0%
1
0%
2
0%
3
2
0%
5
2
Let
a
=
log
3
log
3
2
. An integer k satisfying
1
<
2
(
−
k
+
3
−
a
)
<
2
,
must be less than _____.
Report Question
0%
1.25766
0%
2.256
0%
3
0%
1
If
l
o
g
A
D
=
a
,
then value of
l
o
g
6
12
is (in terms of a)
Report Question
0%
1
+
3
a
3
a
0%
1
+
2
a
3
a
0%
1
+
2
a
2
a
0%
1
+
3
a
2
a
If
x
=
198
!
then value of the expression
1
log
2
x
+
3
log
2
x
+
.
.
.
198
log
2
x
equals ?
Report Question
0%
−
1
0%
0
0%
1
0%
198
Given
l
o
g
2
=
a
,
l
o
g
3
=
b
express the following in terms of
a
or
b
or both
Report Question
0%
log
1.5
0%
log
1.2
0%
log
0.24
0%
log
0.5
0%
log
0.036
If
y
=
a
log
|
x
|
+
b
x
2
+
x
has extreme values at
x
=
2
and
x
=
−
4
/
3
then
Report Question
0%
a
=
12
,
b
=
−
10
0%
a
=
4
,
b
=
−
3
/
4
0%
a
=
−
6
,
b
=
1
/
4
0%
n
o
n
e
The value of
l
o
g
2
24
l
o
g
96
2
−
l
o
g
2
192
l
o
g
12
2
is
Report Question
0%
3
0%
0
0%
2
0%
1
Explanation
Consider
l
o
g
2
24
l
o
g
96
2
−
l
o
g
2
192
l
o
g
12
2
=
l
o
g
24.
l
o
g
96
−
l
o
g
192
l
o
g
12
(
l
o
g
2
)
2
=
l
o
g
(
2
3
×
3
)
l
o
g
(
2
5
×
3
)
−
l
o
g
(
2
6
×
3
)
l
o
g
(
2
2
×
3
)
(
l
o
g
2
)
2
=
(
3
l
o
g
2
+
l
o
g
3
)
(
5
l
o
g
2
+
l
o
g
3
)
−
(
6
l
o
g
2
+
l
o
g
3
)
(
2
l
o
g
2
+
l
o
g
3
)
(
l
o
g
2
)
2
=
15
(
l
o
g
2
)
2
−
12
(
l
o
g
2
)
2
(
l
o
g
2
)
2
=
3
(
l
o
g
2
)
l
o
g
2
=
3
Option A is the correct answer.
The value of
(
0.2
)
l
o
g
√
5
(
1
4
+
1
8
+
1
16
+
.
.
.
)
is
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0%
1
0%
2
0%
1
2
0%
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Answered
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Incorrect : 0
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