Question

Solve!!
$\longmapsto \sf \pink{\frac{({2}^{5})^{2} \times {7}^{3}}{{8}^{3} \times 7}}$
$\longmapsto \sf \pink{\frac{25 \times {5}^{2} \times {t}^{8}}{{10}^{3} \times {t}^{4}}}$


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Answer 1

★ How to do :-

Here, we are given with two fractions which consists exponential numbers in them. We are asked to simplify those equations. These types of numbers are known as base and powers. The number below is the base and the number that is above us the exponent number. These are simplified by a formula. There are six formulas related to exponents and powers. They are also known as Laws of Exponents. They are mentioned below which can be used as a reference. So, let's solve!!

$\:$

➤ Solution (1) :-

${\tt \leadsto \dfrac{({2}^{5})^{2} \times {7}^{3}}{{8}^{3} \times 7}}$

${\tt \leadsto \dfrac{({2}^{5})^{2} \times {7}^{3}}{{({2}^{3})}^{3} \times {7}^{1}}}$

${\tt \leadsto \dfrac{{2}^{5 \times 2} \times {7}^{3}}{{{2}^{3 \times 3}} \times {7}^{1}}}$

${\tt \leadsto \dfrac{{2}^{10} \times {7}^{3}}{{{2}^{9}} \times {7}^{1}}}$

${\tt \leadsto {2}^{10 - 9} \times {7}^{3 - 1}}$

${\tt \leadsto {2}^{1} \times {7}^{2}}$

${\tt \leadsto 2 \times 7 \times 7 = \orange{\boxed{\tt 98}}}$

$\:$

➤ Solution (2) :-

${\tt \leadsto \dfrac{25 \times {5}^{2} \times {t}^{8}}{{{10}^{3}} \times {t}^{4}}}$

${\tt \leadsto \dfrac{{5}^{2} \times {5}^{2} \times {t}^{8}}{{{(2 \times 5)}^{3}} \times {t}^{4}}}$

${\tt \leadsto \dfrac{{5}^{2 + 2} \times {t}^{8}}{{{2}^{3}} \times {5}^{3} \times {t}^{4}}}$

${\tt \leadsto \dfrac{{5}^{4} \times {t}^{8}}{{{2}^{3}} \times {5}^{3} \times {t}^{4}}}$

${\tt \leadsto \dfrac{{5}^{4 - 3} \times {t}^{8 - 4}}{{{2}^{3}}}}$

${\tt \leadsto \dfrac{{5}^{1} \times {t}^{4}}{{{2}^{3}}}}$

${\tt \leadsto \orange{\boxed{\tt \dfrac{{5t}^{4}}{{8}}}}}$

Hence solved !!

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$\dashrightarrow$ Some related equations :-

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

${\sf \longrightarrow {a}^{m} \div {a}^{n} = {a}^{m - n}}$

${\sf \longrightarrow ({a}^{m})^{n} = {a}^{m \times n}}$

${\sf \longrightarrow {a}^{ - n} = \dfrac{1}{{a}^{n}}}$

${\sf \longrightarrow {a}^{0} = 1}$