(sqrt{9})^{3} times (sqrt{81})^{5} div 27^{2} | vedclass

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Question

(D) Given expression: $(\sqrt{9})^{3} 	imes (\sqrt{81})^{5} \div 27^{2} = 3^{x}$Step $1$: Simplify the base values.$\sqrt{9} = 3$,so $(\sqrt{9})^{3}

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Given expression: $(\sqrt{9})^{3} 	imes (\sqrt{81})^{5} \div 27^{2} = 3^{x}$Step $1$: Simplify the base values.$\sqrt{9} = 3$,so $(\sqrt{9})^{3} = 3^{3} = 27$.$\sqrt{81} = 9$,so $(\sqrt{81})^{5} = 9^{5} = (3^{2})^{5} = 3^{10}$.$27^{2} = (3^{3})^{2} = 3^{6}$.Step $2$: Substitute these values into the expression.$3^{3} 	imes 3^{10} \div 3^{6} = 3^{x}$.Step $3$: Apply the laws of exponents ($a^{m} 	imes a^{n} = a^{m+n}$ and $a^{m} \div a^{n} = a^{m-n}$).$3^{3+10-6} = 3^{x}$.$3^{7} = 3^{x}$.Therefore,$x = 7$.

$(\sqrt{9})^{3} \times (\sqrt{81})^{5} \div 27^{2} = 3^{?}$

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