Question

guys Answer this QNo:4​

Answer 1

$\frac{ {9}^{y} . {3}^{2} . {( {3}^{ - y} )}^{ - 1} - {27}^{y} }{ {3}^{3x} . {2}^{3} } = \frac{1}{27} \\ \frac{ {3}^{2y}. {3}^{2} . {3}^{y} - {3}^{3y} }{ {3}^{3x}. {2}^{3} } = \frac{1}{ {3}^{3} } \\ \frac{ {3}^{(2y + y)} . {3}^{2} - {3}^{3y} }{ {3}^{3x} . {2}^{3} } = \frac{1}{ {3}^{3} } \\ \frac{ {3}^{ 3y}. {3}^{2} - {3}^{3y} }{ {3}^{3x} . {2}^{3} } = \frac{1}{ {3}^{3} } \\ \frac{ {3}^{3y}( {3}^{2} - 1)}{ {3}^{3x}.8 } = \frac{1}{ {3}^{3} } \\ \frac{ {3}^{3y}.(9 - 1) }{ {3}^{3x}.8 } = \frac{1}{ {3}^{3} } \\ \frac{ {3}^{3y} .8}{ {3}^{3x} .8 } = \frac{1}{ {3}^{3} } \\ {3}^{(3y - 3x)} = {3}^{ - 3} \\ 3y - 3x = - 3 \\ 3(y - x) = - 3 \\ y - x = - 1 \\ x - y = 1$