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Class 11
Maths
Algebra
Logarithm and Antilogarithm
If
3
1
lo
g
3
M
+
3
lo
g
3
N
=
1
+
lo
g
0
.
0
0
8
5
then
(a)
M
9
=
N
9
(b)
N
9
=
M
9
(c)
M
3
=
N
3
(d)
N
9
=
M
3
Correct answer:
(b)
Solution:
3
1
lo
g
3
M
+
3
lo
g
3
N
=
1
+
lo
g
0
.
0
0
8
5
3
1
lo
g
3
M
+
3
lo
g
3
N
=
1
+
lo
g
0
.
0
0
8
lo
g
5
3
1
lo
g
3
M
+
3
lo
g
3
N
=
1
+
lo
g
1
0
0
0
8
lo
g
5
3
1
lo
g
3
M
+
3
lo
g
3
N
=
1
+
lo
g
8
−
lo
g
1
0
0
0
lo
g
5
3
1
lo
g
3
M
+
3
lo
g
3
N
=
1
+
3
(
lo
g
2
−
lo
g
1
0
)
lo
g
5
3
1
lo
g
3
M
+
3
lo
g
3
N
=
1
+
3
(
lo
g
5
1
lo
g
5
lo
g
3
M
3
1
N
3
=
1
−
3
1
lo
g
M
3
1
N
3
=
3
2
lo
g
3
M
3
1
N
3
=
3
3
2
M
N
9
=
9
N
9
=
M
9
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Similar topics
determinants
matrices
binomial theorem
permutations and combinations
complex numbers and quadratic equations
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View Similar topics ->
determinants
matrices
binomial theorem
permutations and combinations
complex numbers and quadratic equations
View All ↓
If mantissa of logarithm of 719.3 to the base 10 is 0.8569 , then mantissa of logarithm of 71.93 is
The value of
lo
g
1
0
0
.
0
0
1
is equal to
4
3
/
2
=
8
Express each of the following in a form free from logarithm :
lo
g
F
=
lo
g
G
+
lo
g
m
1
+
lo
g
m
2
−
2
lo
g
d
can be expressed as
F
=
G
d
2
m
1
m
2
.
If true write 1 and if false then write 0
Given
3
x
=
9
1
then
x
=
?
Rewrite the following equalities in the exponential form.
l
o
g
1
0
0
0
.
1
=
−
2
1
If
lo
g
y
lo
g
x
=
2
3
and
lo
g
(
x
y
)
=
5
, then number of zeros in
x
×
y
are
Number of real values of
x
satisfying
x
+
lo
g
1
0
(
1
+
2
x
)
=
x
lo
g
1
0
5
+
lo
g
1
0
6
is
If mantissa of logarithm of 719.3 to the base 10 is 0.8569 , then mantissa of logarithm of 71.93 is
The value of
lo
g
1
0
0
.
0
0
1
is equal to
4
3
/
2
=
8
Express each of the following in a form free from logarithm :
lo
g
F
=
lo
g
G
+
lo
g
m
1
+
lo
g
m
2
−
2
lo
g
d
can be expressed as
F
=
G
d
2
m
1
m
2
.
If true write 1 and if false then write 0
Given
3
x
=
9
1
then
x
=
?
Rewrite the following equalities in the exponential form.
l
o
g
1
0
0
0
.
1
=
−
2
1
If
lo
g
y
lo
g
x
=
2
3
and
lo
g
(
x
y
)
=
5
, then number of zeros in
x
×
y
are
Number of real values of
x
satisfying
x
+
lo
g
1
0
(
1
+
2
x
)
=
x
lo
g
1
0
5
+
lo
g
1
0
6
is
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