Question

tv24 3 ki power n upon 5 into 2 ki power 2 X + 1 upon 9 and power and into 3 ki power n -1 ​

Answer 1

Answer:

$\underline{\bigstar\:\:\textsf{According to the Question :}}$

$:\implies\tt \dfrac{(243)^{\frac{n}{5}}\times3^{(2n+1)}}{9^n\times3^{(n-1)}}\\\\\\:\implies\tt \dfrac{(3)^{5 \times \frac{n}{5}}\times3^{(2n+1)}}{3^{2n}\times3^{(n-1)}}\\\\\\:\implies\tt \dfrac{3^n\times3^{(2n+1)}}{3^{2n}\times3^{(n-1)}}\\\\\\:\implies\tt \dfrac{3^{(n + 2n + 1)}}{3^{(2n + n - 1)}} \\\\\\:\implies\tt \dfrac{3^{(3n + 1)}}{3^{(3n - 1)}}\\\\\\:\implies\tt3^{(3n + 1) - (3n - 1)}\\\\\\:\implies\tt3^{(3n + 1 - 3n + 1)}\\\\\\:\implies\tt3^{(1 + 1)}\\\\\\:\implies\tt3^2\\\\\\:\implies\large\underline{\boxed{\textsf{\textbf{9}}}}$

$\rule{150}{1}$

$\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}$

Answer 2

First let's know about law of exponents:-

• $\rm x^a \times x^b = x^{(a + b)}$

• $\rm \dfrac{x^a}{x^b} = x^{(a - b)}$

• $\rm If \: x^a = x^b \: then, a = b$

• $\rm \dfrac{1}{a} = a^{-1}$

• $\rm (x^a)^b = x^{ab}$

Solution:-

Given to evaluate:-

$\star \quad\sf \dfrac{ (243)^{\frac{n}{5}} \times 3^{2n + 1}}{9^n \times 3^{n - 1}}\\\\\\\quad\dashrightarrow\sf \dfrac{ (3^5)^{\frac{n}{5}} \times 3^{2n + 1}}{(3^2)^n \times 3^{n - 1}}\\\\\\\quad\dashrightarrow\sf \dfrac{3^{\frac{5n}{5}} \times 3^{2n + 1}}{3^{2n} \times 3^{n - 1}}\\\\\\\quad\dashrightarrow\sf \dfrac{3^n \times 3^{2n + 1}}{3^{2n} \times 3^{n - 1}} \\\\\\\quad\dashrightarrow\sf \dfrac{3^{(n + 2n + 1)}}{3^{(2n + n - 1)}}\\\\\\\quad\dashrightarrow\sf \dfrac{3^{(3n + 1)}}{3^{(3n - 1)}}\\\\\\\quad\dashrightarrow\sf 3^{(3n + 1 - 3n + 1)}\\\\\\\quad\dashrightarrow\sf 3^2 \\\\\\\quad\dashrightarrow\large\boxed{\boxed{\sf\blue{\: 9 \: }}}$

Therefore,

Required value = 9