Question

evaluate(3-⁵×10-⁵×5³)÷(5-⁷×6-⁵×5⁵)​

Answer 1

$\sf\huge\bold{Answer:}$

Given :

$⇒( {3}^{ - 5} \times {10}^{ - 5} \times {5}^{3} ) \div ( {5}^{ - 7} \times {5}^{5} )$

Solution :

$⇒( {3}^{ - 5} \times {10}^{ - 5} \times {5}^{3} ) \div ( {5}^{ - 7} \times {5}^{5} )$

we know that,

$⇒a \div b = a \times \frac{1}{b} = \frac{a}{b}$

similarly,using the above identity in the term,

we get

$=( {3}^{ - 5} \times {10}^{ - 5} \times {5}^{3} ) \times \frac{1}{( {5}^{ - 7} \times {5}^{5} )} \\ = \frac{( {3}^{ - 5} \times {10}^{ - 5} \times {5}^{3} )}{ {5}^{ - 7} \times {5}^{5} }$

now,using identities

$⇒{a}^{m} \times {b}^{m} = {(ab)}^{m} \\ ⇒ {a}^{m} \times {a}^{n} = {a}^{m + n} \\ ⇒ {a}^{m} \div {a}^{n} = {a}^{m - n}$

$= \frac{ {(3 \times 10) }^{ - 5} \times {5}^{3} }{ {5}^{ - 7 + 5} }$

Now,we also know that

$⇒ {a}^{ m} = \frac{1}{ {a}^{ - m} } \: or \: vice \: - versa$

so,using this in the given term,

we get

$= \frac{ {30}^{ - 5} }{ {5}^{ - 2} \times {5}^{ - 3} } \\ = \frac{ {30}^{ - 5} }{ {5}^{ - 2 - 3} } \\ = \frac{ {30}^{ - 5} }{ {5}^{ - 5} }$

so,in the above identities the reciprocal of those identities is also true

$⇒ {ab}^{m} = {a}^{m} \times {b}^{m}$

Using this,we get

$( {6 \times 5)}^{ - 5} on \: splitting \: {30}^{ - 5}$

$= \frac{ {6}^{ - 5} \times {5}^{ - 5} }{ {5}^{ - 5} } \\ = {6}^{ - 5} \: or \: \frac{1}{ {6}^{5} }$

so,the final solution after step by step explanation of $⇒( {3}^{ - 5} \times {10}^{ - 5} \times {5}^{3} ) \div ( {5}^{ - 7} \times {5}^{5} )$ is ${6}^{-5}$