9^{3} times 81^{2} div 27^{3} = (3)^{?} | vedclass

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(C) Given expression: $9^{3} 	imes 81^{2} \div 27^{3} = (3)^{?}$Express each base as a power of $3$:$9 = 3^{2}$,$81 = 3^{4}$,and $27 = 3^{3}$.Substit

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given expression: $9^{3} 	imes 81^{2} \div 27^{3} = (3)^{?}$Express each base as a power of $3$:$9 = 3^{2}$,$81 = 3^{4}$,and $27 = 3^{3}$.Substituting these values into the expression:$(3^{2})^{3} 	imes (3^{4})^{2} \div (3^{3})^{3} = (3)^{?}$Using the power of a power rule $(a^{m})^{n} = a^{m 	imes n}$:$3^{6} 	imes 3^{8} \div 3^{9} = (3)^{?}$Using the product and quotient rules for exponents $a^{m} 	imes a^{n} = a^{m+n}$ and $a^{m} \div a^{n} = a^{m-n}$:$3^{6+8-9} = (3)^{?}$$3^{5} = (3)^{?}$Comparing the exponents,we get $? = 5$.

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

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