Question

pls can anyone solve this sum chap- Indices​

Answer 1

Answer: m = 1+ n

$\frac{ {3}^{2n} . {3}^{2} . {3}^{n} - {3}^{3n} }{ {3}^{3m} . {2}^{3} } = \frac{1}{ {3}^{3} } \\ \\ \frac{ {3}^{3n + 2} - {3}^{3n} }{ {3}^{3m} } = \frac{ {2}^{3} }{ {3}^{3} } \\ \\ \frac{ {3}^{3n}( {3}^{2 } - 1) }{ {3}^{3m} } = {( \frac{2}{3} )}^{3} \\ \\ \\ \\ \frac{8}{ {3}^{3m - 3n} } = {( \frac{2}{3} )}^{3} \\ \\ \\ \\ {( \frac{1}{ {3}^{m - n} } )}^{3} = {( \frac{1}{3} )}^{3} \\ \\ {3}^{m - n} = {3}^{1} \\ \\ m - n = 1 \\ \\ m = 1 + n$