A) 2
B) 65/8
C) 37/6
D) None of these
Correct Answer: D
Solution :
We have \[{{\log }_{2}}x+\frac{1}{{{\log }_{2}}x}=3+\frac{1}{3}={{\log }_{2}}y+\frac{1}{{{\log }_{2}}y}\] \ \[{{\log }_{2}}x=3,{{\log }_{2}}y=\frac{1}{3}\] \[(\because x\ne y)\] Þ \[x={{2}^{3}}\]and\[y={{2}^{1/3}}\Rightarrow x+y=8+{{2}^{1/3}}\].You need to login to perform this action.
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