Question

The valu of (243)^n/5*3^2n+1/9n*3^n-1​

Answer 1

Answer:

First converting 243 in terms of 3

3^5n/5*3^2n+1/9n*3^n-1

(when bases are same with * sign we can add the powers)

3^n+2n+1/9n+n-1

(taking lcm)

3^9n^2+2n+1+9n^2-9n

3^18n^2-7n+1

3^18n^2-9n-2n+1

3^9n(2n-1)-1(2n-1)

3^(9n-1)(2n-1)        [equating both the brackets with 0]

So, we will get 2values of n=1/9 and1/2

value of the given question will be 3^1/9 and 3^1/2.

Hope it helps you....

Answer 2

$\huge{\boxed{\boxed{\bf{\green{Question}}}}}$

$\large{\bf{The\:value\:of\: \frac{{243}^{ \frac{n}{5} } \times {3}^{2n + 1} }{ {9}^{n} \times {3}^{n - 1} } }}$

$\huge{\boxed{\boxed{\bf{\orange{Answer}}}}}$

$\frac{{243}^{ \frac{n}{5} } \times {3}^{2n + 1} }{ {9}^{n} \times {3}^{n - 1} } \\ \\ \\ = > \frac{{ ({3}^{5}) }^{ \frac{n}{5} } \times {3}^{2n + 1} }{ {( {3}^{2}) }^{n} \times {3}^{(n - 1)} } \\ \\ \\ = > \frac{ {3}^{(5 \times \frac{n}{5}) \times {3}^{(2n + 1)} } }{ {3}^{2n} \times {3}^{(n - 1)} } \\ \\ \\ = > \frac{ {3}^{(n + 2n + 1)} }{ {3}^{(2n + n - 1)} } \\ \\ \\ = > \frac{{3}^{(3n + 1)} }{ {3}^{(3n - 1)} } \\ \\ \\ = > {3}^{(3n + 1) - (3n - 1)} \\ \\ = > {3}^{2} = 9$

$\large{\bf{\pink{Hence,\:the\: value\:of\: given\: question\:is\:9}}}$