The value of frac{(243)^{frac{n}{5}} times 3^ | vedclass

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(D) Given expression: $\frac{(243)^{\frac{n}{5}} 	imes 3^{2n+1}}{9^{n} 	imes 3^{n-1}}$Express all terms with base $3$:$243 = 3^5$ and $9 = 3^2$.Subs

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$9$

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(D) Given expression: $\frac{(243)^{\frac{n}{5}} 	imes 3^{2n+1}}{9^{n} 	imes 3^{n-1}}$Express all terms with base $3$:$243 = 3^5$ and $9 = 3^2$.Substitute these into the expression:$\frac{(3^5)^{\frac{n}{5}} 	imes 3^{2n+1}}{(3^2)^n 	imes 3^{n-1}}$Simplify the exponents:$= \frac{3^n 	imes 3^{2n+1}}{3^{2n} 	imes 3^{n-1}}$Combine the exponents using the rule $a^m 	imes a^n = a^{m+n}$:$= \frac{3^{n + 2n + 1}}{3^{2n + n - 1}} = \frac{3^{3n+1}}{3^{3n-1}}$Use the rule $\frac{a^m}{a^n} = a^{m-n}$:$= 3^{(3n+1) - (3n-1)} = 3^{3n+1-3n+1} = 3^2 = 9$.

The value of $\frac{(243)^{\frac{n}{5}} \times 3^{2n+1}}{9^{n} \times 3^{n-1}}$ is

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