Question

Please solve this both question. I will mark one brainlist who will solve correctly

Answer 1

Answer:

answer is B and C.........

Answer 2

Here's Your answer Dude :)

1.

$( - 2) ^{ - 5} \div \frac{2 ^{ - 3} \times 3 ^{7} }{3 ^{ - 6} \times 2 ^{ - 5} }$

So, First Reshuffle it..

$( - 2) ^{ - 5} \div \frac{2 ^{ - 3} \times 3 ^{7} }{2 ^{ - 5} \times 3 ^{ - 6} }$

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you must be knowing the laws of exponent i.e.. 《if the bases are same while dividing the powers are subtracted 》

$\frac{a ^{m} }{a ^{n} } = a ^{m - n}$

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Now after using the law we have...

$= > ( - 2) ^{ - 5} \div 2 ^{ - 3 - ( - 5)} \times 3 ^{7 - ( - 6)}$

$= > ( - 2) ^{ - 5} \div 2 ^{ - 3 + 5} \times 3 ^{7 + 6}$

$= > ( - 2) ^{ - 5} \div 2 ^{2} \times 3 ^{13}$

$= > \frac{( - 2) ^{ - 5} }{2 ^{2} \times 3 ^{13} }$

Power negative so reciprocal:

$\frac{1}{( - 2) ^{ 5} \times 2 ^{2} \times 3 ^{13} }$

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Law of exponent says:

If base is same while multiplying we may add the powers

$a ^{m} \times a^{n} = a ^{m + n}$

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After using the law we have :

$= > \frac{ - 1}{2 ^{5} \: \times 2 ^{2} \: \times \: 3 ^{13} }$

$= > \frac{ - 1}{ {2}^{5 + 2} \: \times \: 3 ^{13} }$

$= > \frac{ - 1}{2 ^{7} \: \times \: 3 ^{13} }$

(1st) option is correct

FIRST QUESTION DONE

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2.

$\sqrt{7 \sqrt{7 \sqrt{7 \sqrt{7 \sqrt{7} } } } } = 7 ^{x}$

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$\sqrt{x} = x ^{ \frac{1}{2} }$

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So after using the above rule we have:

$= > 7 ^{ \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} } = 7 ^{x}$

$= > 7 ^{ \frac{1}{32} } = 7 ^{x}$

Equating the powers :

$x = \frac{1}{32}$

(4th) option is correct

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Hope u Understand :)

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