Question

Pls Solve 2, 3 , 5 , 6, 9, 10, 11 sums ​

Answer 1

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⧠ Question:-

( (5²)³ × 5⁴ ) ÷ 5⁷

Law of Exponents to be used:

$~~~~~~\longrightarrow \large[ ( \sf {a}^{m} ) \sf {}^{n} = \sf{a}^{mn} ]$

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

$~~~~~~\longrightarrow \large\sf ( {a}^{m} \div {a}^{n} = {a}^{m - n})$

Solution:-

$\begin{gathered}\large : \implies[ ( \sf {a}^{m} ) \sf {}^{n} = \sf{a}^{mn} ] \\\\\\ \longrightarrow \sf [ ( {5}^{2 \times 3 } ) \times {5}^{4} \div {5}^{7} ] \\\\\\\longrightarrow \sf [ ( {5}^{6} \times {5}^{4} ) \div {5}^{7} \\\\\\ \dag{\underline {\frak{\pink{~as~we~know~that : \: 2nd \: step \: }}}} \end{gathered}$

$\begin{gathered} \longrightarrow \sf \large ( {a}^{m} \times {a}^{n} = {a}^{ m + n })\\\\\\ \longrightarrow \sf ( {5}^{6 + 4 } )\div {5}^{7} \\\\\\ \longrightarrow \sf {5}^{10} \div {5}^{7} \\ \end{gathered}$

${\dag\: {\underline {\frak{\orange{ now, ~3rd ~step~we~know~that}}}}}$

$\begin{gathered}\sf\large~~~~~ \longrightarrow \: ( {a}^{m} \div {a}^{n} = {a}^{m - n}) \\\\\\ \longrightarrow \sf {5}^{10-7} \\\\\\ \sf = {5}^{3}\end{gathered}$

$\therefore\large \frak{\underline {\color{pink}{~solution ~of~2nd~sum~is \: \: {5}^{3}}}}$

⠀⠀

⠀⠀

Question :

25⁴ ÷ 5³

Solution:

⠀⠀⠀⠀[ Refer Attachment ] = 5⁵ ✔

⠀⠀

Question:

$\sf \dfrac{ {3}^{7} }{ {3}^{4} \times {3}^{3} }$

Solution:-

⠀⠀⠀⠀1 ( Refer Attachment [2] ) ✔

⠀⠀

Question:

2⁰ + 3⁰ + 4⁰

Law of Exponents to be used:

$\sf\orange { ( { {a}^{0} = 1} )}$

Solution:-

⠀⠀⠀⠀2⁰ = 1 , 3⁰ = 1 , 4⁰ = 1

⠀⠀So,

⠀⠀⠀⠀⠀⠀1 + 1 + 1 = 3

Question:-

$\sf\dfrac{ {2}^{8} \times {a}^{5} }{ {4}^{3} \times {a}^{3} }$

Solution:-

⠀⠀⠀⠀[ Refer Attachment 3 ] ✔ = 2² - a²

Question 10:

$\bigg ( \sf\dfrac{ {a}^{5} }{ {a}^{3} } \bigg ) \: \times {a}^{8}$

Solution:-

⠀⠀⠀⠀[ Refer Attachment 4 ] ✔ = a¹⁰

Question 11:

$\sf \dfrac{ {4}^{5} \times {a}^{8} {b}^{3} }{ {4}^{5} \times {a}^{5} {b}^{2} }$

Solution:

⠀⠀⠀⠀[ Refer Attachment 4 ] ✔ = a³b

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$\begin{gathered}\begin{gathered} \small\boxed{ \begin{array}{cc}\Large \underline{\textsf{\textbf{More law \: of \: exponents}}} \\ \\ \\ \: \: \: \: \: \: \bigstar \: \: \: \large\sf{ \:({a}^{m} \times {a}^{n} = {a}^{m+n} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \\ \bigstar~ \:\large \sf{ \: ({a}^{m} \div {a}^{n} = {a}^{m-n} ) } \\ \\ \\ \bigstar \: \large\sf{ \:[( {a}^{m} ){}^{n} = {a}^{m-n}] } \\ \\ \\ \bigstar \: \large\sf{ \: \bigg(\dfrac { {a}^{m} }{ {a}^{n} } \bigg) = {a}^{m - n} } \\ \\ \\ \bigstar \: \large\sf{{a}^{m} \times {b}^{m} = ( {ab} ) {}^{m}} \\ \\ \\ \bigstar \: \large\sf{ \dfrac{ {a}^{m} }{ {b}^{m} } = \bigg( \dfrac{a}{b}\bigg) {}^{m}} \\ \\ \\\bigstar \: \large\sf{ {a}^{0} = 1} \\ \\ \\ \end{array} }\end{gathered}\end{gathered}$

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