Question
If 3x+y = 81 and 81x-y = 3, then the value of $$\frac{x}{y}$$ is = ?
Answer: Option D
$$\eqalign{
& {\text{According to question,}} \cr
& \Rightarrow {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr
& \Rightarrow {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr
& \Rightarrow x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr
& x + y = 4......{\text{(i)}} \cr
& {\text{ }}x - y = \frac{1}{4}.....(ii) \cr
& {\text{Solve the equation of (i) and (ii)}} \cr
& x = \frac{{17}}{8}, \cr
& y = \frac{{15}}{8}, \cr
& \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$
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$$\eqalign{
& {\text{According to question,}} \cr
& \Rightarrow {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr
& \Rightarrow {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr
& \Rightarrow x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr
& x + y = 4......{\text{(i)}} \cr
& {\text{ }}x - y = \frac{1}{4}.....(ii) \cr
& {\text{Solve the equation of (i) and (ii)}} \cr
& x = \frac{{17}}{8}, \cr
& y = \frac{{15}}{8}, \cr
& \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$
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