Answer:
\[\frac{{{9}^{n}}\times {{3}^{2}}\times {{3}^{n}}-{{(27)}^{n}}}{{{\left( {{3}^{3}} \right)}^{5}}\times {{2}^{3}}}=\frac{1}{27}\] \[\frac{{{\left( 3\times 3 \right)}^{n}}\times {{3}^{2}}\times {{3}^{n}}-{{(3\times 3\times 3)}^{n}}}{{{3}^{15}}\times {{2}^{3}}}=\frac{1}{3\times 3\times 3}\] \[\frac{{{\left( {{3}^{2}} \right)}^{n}}\times {{3}^{n+2}}-{{\left( {{3}^{3}} \right)}^{n}}}{{{3}^{15}}\times {{2}^{3}}}=\frac{1}{{{3}^{3}}}\] \[\frac{{{3}^{2}}^{n}\times {{3}^{n+2}}-{{3}^{3}}^{n}}{{{3}^{15}}\times {{2}^{3}}}={{3}^{-3}}\] \[\frac{{{3}^{2}}{{^{n}}^{+n+2}}-{{3}^{3}}^{n}}{{{3}^{15}}\times {{2}^{3}}}={{3}^{-3}}\] \[\frac{{{3}^{2}}{{^{n}}^{+n+2}}-{{3}^{3}}^{n}}{{{3}^{15}}\times {{2}^{3}}}={{3}^{-3}}\] \[\frac{{{3}^{3n}}({{3}^{2}}-1)}{{{3}^{15}}\times {{2}^{3}}}={{3}^{-3}}\] \[\frac{{{3}^{3n-15}}(9-1)}{{{2}^{3}}}={{3}^{-3}}\] \[{{3}^{3n-15}}\times \frac{8}{8}={{3}^{-3}}\] \[{{3}^{3n-15}}={{3}^{-3}}\] As base are same on both sides, so 3n - 15 = - 3 3n = - 3 + 15 3n = 12 thus, n = \[\frac{12}{3}\] = 4
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