A) 3 and 12
B) 12 and 3
C) 6 and 6
D) 4 and 9
Correct Answer: A
Solution :
\[{{27}^{x+1}}={{9}^{x+3}}\] \[\Rightarrow \] \[{{({{3}^{3}})}^{x+1}}={{({{3}^{2}})}^{x+3}}\] \[\Rightarrow \] \[{{3}^{3x+3}}={{3}^{2x}}+6\] \[\Rightarrow \] \[3x+3=2x+6\] \[\Rightarrow \] \[3x-2x=6-3\] \[\Rightarrow \] \[x=3\] Consider \[{{9}^{x+3}}={{3}^{y}}\] \[\Rightarrow \] \[{{({{3}^{2}})}^{x+3}}={{3}^{y}}\] \[\Rightarrow \] \[{{3}^{2x+6}}={{3}^{y}}\] \[\Rightarrow \] \[2x+6=y\] \[\therefore \] \[2(3)+6=y\] \[\Rightarrow \] \[y=12\]
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