Question

9^n*3^2*3^n-(27)^n /(3^3)^5*2^3=1/27

Answer 1

(9^n*3^2*3^n-(27)^n)/(3^3)^5*2^3=1/27

(3^(2n)*3^2*3^n-(27)^n)/(3^(15)*2^3)=1/27

(3^(3n+2)-3^(3n))/(3^(15)*2^3)=1/3^3

3^(3n+5)-3^(3n+3)=3^(15)*2^3

3^(3n+3)(3^2-1)=3^(15)*8

3^(3n+3)*8=3^(15)*8

Therefore, 3n+3=15 => n=4

Answer 2

$\huge{\underline{\underline{\bf{\purple{Solution:}}}}}$

$\frac{ {9}^{n} \times {3}^{2} \times {3}^{n} - {(27)}^{n} }{ {3}^{(3 \times 5)} \times {2}^{3} } = \frac{1}{27}$

$\frac{ {9}^{n} \times {3}^{2} \times \ {3}^{n} - {(27)}^{n} }{ {3}^{(3 \times 5)} \times {2}^{3} } = \frac{1}{ ({3}^{3}) }$

$= \frac{ {3}^{2n} \times {3}^{2} \times {3}^{n} - {(27)}^{n} }{ {3}^{15} \times {2}^{3} }$

$= \frac{ {3}^{2n} \times {3}^{2} \times {3}^{n} - {(27)}^{n} }{ {3}^{15} \times {2}^{3} = {3}^{-3} }$

$= [{3}^{(2n+n+2)} - {3}^{3n}] = {3}^{-3} \times {3}^{15} \times{2}^{3} }$

$= {3}^{(3n+2)} - {3}^{3n} = {3}^{-3+15} \times {2}^{3} }$

$= {3}^{3n} ({3}^{2} - 1) = {3}^{12} \times {2}^{3} }$

$= {3}^{3n} \times 8 = {3}^{12} \times 8 } }$

$= {3}^{3n} = \frac{ {3}^{12} \times 8}{8} = {3}^{12} }$

$\huge{\underline{\underline{\bf{\orange{3n=12}}}}}$

$\huge{\underline{\underline{\bf{\green{n=4}}}}}$

Hence, n = 4