Question
1417×1417 can be represented as ___________.
Answer: Option D
:
D
For any positive real number 'a' and integers 'm' and 'n', we define
1.am×an=am+n
2. (am)n=amn
3. (ab)m=ambm andifa=b, the expression becomes (a×a)m=(a2)m=a2m.
So, using first rule, we have
1417+17=1427
Now, using secondrule,
(142)17= 1427
Finally, using thirdrule,
(14×14)17=1427
Hence, we see that all the three results are same.
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:
D
For any positive real number 'a' and integers 'm' and 'n', we define
1.am×an=am+n
2. (am)n=amn
3. (ab)m=ambm andifa=b, the expression becomes (a×a)m=(a2)m=a2m.
So, using first rule, we have
1417+17=1427
Now, using secondrule,
(142)17= 1427
Finally, using thirdrule,
(14×14)17=1427
Hence, we see that all the three results are same.
Was this answer helpful ?
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