A) no solution
B) unique solution
C) infinitely many solutions
D) \[x=2\frac{1}{8},\,y=1\frac{7}{8}\]
Correct Answer: D
Solution :
Given, |
\[{{3}^{x+y}}=81\Rightarrow {{3}^{x+y}}={{3}^{4}}\Rightarrow x+y=4\] (i) |
And \[{{81}^{x-y}}=3\Rightarrow {{3}^{4\left( x-y \right)}}={{3}^{1}}\Rightarrow 4\left( x-y \right)=1\] |
\[\Rightarrow \,x-y=\frac{1}{4}\] (ii) |
On adding Eqs. (i) and (ii), we get |
\[2x=4+\frac{1}{4}=\frac{17}{4}\] |
\[\Rightarrow \,\,\,x=\frac{17}{8}=2\frac{1}{8}\] |
From Eq. (i), we get |
\[y=\frac{15}{8}=1\frac{7}{8}\] |
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