A) \[\frac{1}{3}\]
B) \[\frac{1}{9}\]
C) 3
D) 1
Correct Answer: D
Solution :
(d): \[~x=\text{lo}{{\text{g}}_{3}}27~~~~~y=\text{lo}{{\text{g}}_{9}}27\] \[\Rightarrow x=\text{lo}{{\text{g}}_{3}}{{3}^{3}}\,\,\,\,y=\text{lo}{{\text{g}}_{{{3}^{2}}}}{{3}^{3}}\] \[\Rightarrow x=3~~~~~~y=\frac{3}{2}\] \[\therefore \frac{1}{x}+\frac{1}{y}=\frac{1}{3}+\frac{1}{\left( \frac{3}{2} \right)}=\frac{1}{3}+\frac{2}{3}=1.\]
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