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Standard VII
Mathematics
Exponents with Like Bases
Question
Solve the following using Product Law of Exponents.
a
×
a
2
×
a
1
2
a
7
a
2
7
a
3
a
7
2
A
a
7
B
a
2
7
C
a
7
2
D
a
3
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Solution
Verified by Toppr
By product law of exponents
a
×
a
2
×
a
1
2
=
a
(
1
+
2
+
1
2
)
=
a
7
2
Therefore option
D
is the correct answer.
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Similar Questions
Q1
Solve the following using Product Law of Exponents.
a
×
a
2
×
a
1
2
View Solution
Q2
Suppose
a
2
,
a
3
,
a
4
,
a
5
,
a
6
,
a
7
are integers such that
5
7
=
a
2
2
!
+
a
3
3
!
+
a
4
4
+
a
5
5
!
+
a
6
6
!
+
a
7
7
!
where
0
≤
a
<
j
for
j
=
2
,
4
,
5
,
6
,
7.
The sum
a
2
+
a
3
+
a
4
+
a
5
+
a
6
+
a
7
is
View Solution
Q3
For an A.P.,
a
1
,
a
2
,
a
3
,
…
…
, if
a
3
+
a
5
+
a
8
=
11
and
a
4
+
a
2
=
−
2
, then the value of
a
1
+
a
6
+
a
7
is
View Solution
Q4
Suppose
a
2
,
a
3
,
a
4
,
a
5
,
a
6
,
a
7
are intagers such that
5
7
=
a
2
2
!
+
a
3
3
!
+
a
4
4
!
+
a
5
5
!
+
a
6
6
!
+
a
7
7
!
where
0
≤
a
j
<
j
for
j
=
2
,
3
,
4
,
5
,
6
,
7
. The sum
a
2
+
a
3
+
a
4
+
a
5
+
a
6
+
a
7
is-
View Solution
Q5
For
A
P
a
1
,
a
2
,
a
3
,
..., if
a
4
a
7
=
2
3
, find
a
6
a
8
View Solution