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Byju's Answer
Standard VII
Mathematics
Positive Exponents
Find:i 641/2 ...
Question
Find:
(i)
64
1
2
.
(ii)
32
1
5
.
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Solution
(i)
64
1
2
=
(
2
6
)
1
2
[
∵
64
=
2
6
]
=
2
6
×
1
2
=
2
3
=
8
[
∵
(
a
m
)
n
=
a
m
n
]
(ii)
32
1
5
=
(
2
5
)
1
5
[
∵
32
=
2
5
]
=
(
2
)
5
×
1
5
=
2
[
∵
(
a
m
)
n
=
a
m
n
]
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