A) 42
B) \[\frac{15}{8}\]
C) \[\frac{17}{8}\]
D) 39
Correct Answer: C
Solution :
\[{{3}^{x\,+\,y}}=81\] \[\Rightarrow \] \[{{3}^{x\,+\,y}}={{3}^{4}}\] \[\Rightarrow \] \[x+y=4\] ? (i) and \[{{81}^{x\,+\,y}}=3\] \[\Rightarrow \] \[{{({{3}^{4}})}^{x\,-\,y}}=3\] \[\Rightarrow \] \[{{3}^{4x\,-\,4y}}={{3}^{1}}\] \[\Rightarrow \] \[4x-4y=1\] ? (ii) By (i) \[\times \,4+\] Eq. (ii), we have \[\begin{align} & 4x+4y=16 \\ & \underline{4x-4y=\,\,\,\,1} \\ & 8x=17\Rightarrow x=\frac{17}{8} \\ \end{align}\]
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