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(a^m)^n=a^mn and a^0 = 1

Question 62In...

Question

Question 62

In the following question, fill in the blanks to make the statements true.

By solving $(6^{0} - 7^{0}) \times (6^{0} + 7^{0})$ , we get ___ .

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Solution

Using law of exponents $a^{0} = 1$
$∴ (6^{0} - 7^{0}) \times (6^{0} + 7^{0}) = (1 - 1) \times (1 + 1) = 0 \times 2 = 0$
Hence $(6^{0} - 7^{0}) \times (6^{0} + 7^{0}) = 0$

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(a^m)^n=a^mn and a^0 = 1

Standard VI Mathematics

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