If 3^x+y = 81 and 81^x-y = 3, then the value of $$\frac{x}{y}$$ is = ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{According to question,}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr & \Rightarrow x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr & x + y = 4......{\text{(i)}} \cr & {\text{ }}x - y = \frac{1}{4}.....(ii) \cr & {\text{Solve the equation of (i) and (ii)}} \cr & x = \frac{{17}}{8}, \cr & y = \frac{{15}}{8}, \cr & \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$