A) 0
B) 1
C) \[\frac{3}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: A
Solution :
[a] \[{{2}^{x}}={{3}^{y}}={{6}^{-z}}=k\,\,(say)\] \[\Rightarrow \,2={{k}^{\frac{1}{x}}};3={{k}^{\frac{1}{y}}}\,\,and\,\,6={{k}^{-\frac{1}{z}}}\] \[6=2\times 3={{k}^{\frac{1}{x}}}.{{k}^{\frac{1}{y}}}={{k}^{\frac{1}{x}+\frac{1}{y}}}\] \[\Rightarrow k\left( \frac{1}{x}+\frac{1}{y} \right)={{k}^{-\frac{1}{z}}}\] \[\therefore \,\,\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\,\,or\,\,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.\]
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