Question

4⁵ *a⁷*b⁵/4²*a⁵*b²=
​

Answer 1

Answer :-

$\implies\sf \dfrac{4^5 \times a^7 \times b^5}{4^2 \times a^5 \times b^2}$

$\implies\sf \dfrac{4^5}{4^2} \times \dfrac{a^7}{a^5} \times \dfrac{b^5}{b^2}$

Using the property -

• $\sf \dfrac{a^m}{a^n} = a^{m-n}$

$\implies\sf 4^{5-2} \times a^{7-5} \times b^{5-2}$

$\implies\sf 4^3 \times a^2 \times b^3$

• $\sf 4^3 = 4 \times 4 \times 4 = 64$

$\implies\sf 64a^2b^3$

$\boxed{\sf \dfrac{4^5 \times a^7 \times b^5}{4^2 \times a^5 \times b^2} = 64a^2b^3}$

Additional information :-

$\underline{\text{Law of Exponents :}}$

$\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}$

$\bigstar\:\:\sf{(a^m)^n = a^{mn}}$

$\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}$

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

$\bigstar\:\:\sf\sqrt[ n]{\sf a} = (a)^{\dfrac{1}{n}}$

Answer 2

Given : Expression = $\longmapsto \bf \dfrac{4^5 \times a^{7} \times b^5 }{4^2 \times a^5 \times b^2}$

Exigency To Solve : $\longmapsto \bf \dfrac{4^5 \times a^{7} \times b^5 }{4^2 \times a^5 \times b^2}$

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$\qquad \qquad \bf{\large{\dag}}\;\;\bf \dfrac{4^5 \times a^{7} \times b^5 }{4^2 \times a^5 \times b^2 }$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Solving \: the \: Given \: Expression \::}}$

$\qquad:\implies \sf \dfrac{4^5 \times a^{7} \times b^5 }{4^2 \times a^5 \times b^2}$

Or ,

$\qquad:\implies \sf \dfrac{4^5 }{4^2}\times \dfrac{a^{7}}{a^5} \times \dfrac{b^5}{b^2 }$

$\dag\:\it{As,\:We\:know\:that\::}$

• $\dfrac{a^m}{a^n} = a^{m-n}$

$\qquad:\implies \sf 4^{5-2}\times a^{7-5} \times b^{5-2}$

$\qquad:\implies \sf 4^{3}\times a^{2} \times b^{3}$

$\qquad:\implies \sf 4\times 4\times 4 \times a^{2} \times b^{3}$

$\qquad:\implies \sf 16 \times 4 \times a^{2} \times b^{3}$

$\qquad:\implies \sf 64\times a^{2} \times b^{3}$

Or ,

$\qquad:\implies \bf 64a^{2}b^{3} \:\:\longrightarrow \:Required\:AnswEr \:$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{array}}$

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