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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
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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
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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
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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
∴
\therefore 3px = -24x
\therefore p = -8
Given
log_3(a) = c
and
log_3(b)=2c, a =
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0%
3c
0%
c + 3
0%
b^2
0%
\sqrt{b}
0%
\dfrac{b}{2}
Explanation
Given,
\log _{ 3 }{ (a) } =c
and
\log _{ 3 }{ (b) } =2c
\therefore a={ 3 }^{ c }
and
b={ 3 }^{ 2c }
Now consider
a^2
as follows:
a^{ 2 }={ (3 }^{ c })^{ 2 }
\Rightarrow a^{ 2 }={ 3 }^{ 2c }
\Rightarrow a^{ 2 }=b
\Rightarrow a=\sqrt { 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}+log_{10}(31)
=\log_{10}(2^{6}.31)
=\log_{10}(64\times31)
=\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
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0%
18
0%
24
0%
22
0%
20
Explanation
Given
\log _{ 10 }{ 2 } =0.301
\log _{ 10 }{ 2^{64} } =64 \times \log _{ 10 }{ 2 } =64 \times 0.3010=19.264
\Rightarrow 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
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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\times 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
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0%
1
0%
2
0%
\frac{3}{2}
0%
\frac{5}{2}
Let
a = \log_3\log_32
. An integer k satisfying
1< 2^{(-k+3^{-a})} < 2,
must be less than _____.
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0%
1.25766
0%
2.256
0%
3
0%
1
If
log_AD= a,
then value of
log_612
is (in terms of a)
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0%
\frac{1+3a}{3a}
0%
\frac{1+2a}{3a}
0%
\frac{1+2a}{2a}
0%
\frac{1+3a}{2a}
If
x=198!
then value of the expression
\dfrac {1}{\log_{2}x}+\dfrac {3}{\log_{2}x}+...\dfrac {198}{\log_{2}x}
equals ?
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0%
-1
0%
0
0%
1
0%
198
Given
log2=a,log3=b
express the following in terms of
a
or
b
or both
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0%
\log1.5
0%
\log1.2
0%
\log0.24
0%
\log0.5
0%
\log0.036
If
y=a\log\left|x\right|+bx^{2}+x
has extreme values at
x=2
and
x=-4/3
then
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0%
a=12,b=-10
0%
a=4,b=-3/4
0%
a=-6,b=1/4
0%
none
The value of
\dfrac{log_2 24}{log_{96} 2}-\dfrac{log_2192}{log_{12}{2}}
is
Report Question
0%
3
0%
0
0%
2
0%
1
Explanation
Consider
\dfrac{log_2 24}{log_{96} 2}-\dfrac{log_2192}{log_{12}{2}}\\
=\dfrac{log24.log96-log192log12}{(log2)^2}
=\dfrac{log(2^3 \times 3)log(2^5\times 3)-log(2^6\times3)log(2^2\times3)}{(log2)^2}
=\dfrac{(3log2+log3)(5log2+log3)-(6log2+log3)(2log2+log3)}{(log2)^2}
=\dfrac{15(log2)^2-12(log2)^2}{(log2)^2}
=3\dfrac{(log2)}{log2}
=3
Option A is the correct answer.
The value of
(0.2)^{log_{\sqrt{5}} \left(\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + ...\right)}
is
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0%
1
0%
2
0%
\dfrac{1}{2}
0%
4
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