| If \[{{2}^{x}}={{3}^{y}}={{6}^{-\,\,z}},\]then \[\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\]is equal to |
A) \[0\]
B) \[1\]
C) \[\frac{3}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: A
Solution :
| Let \[{{2}^{x}}={{3}^{y}}={{6}^{-z}}=k\] |
| We know that, If \[{{a}^{x}}=y,\] then \[a={{y}^{1/x}}\] |
| \[\Rightarrow \] \[2={{k}^{\frac{1}{x}}};3={{k}^{\frac{1}{y}}};6={{k}^{-\frac{1}{z}}}\] |
| \[\Rightarrow \] \[{{k}^{\frac{1}{x}}}\times {{k}^{\frac{1}{y}}}={{k}^{-\,\,\frac{1}{z}}}\] \[[\because 2\ \times 3=6]\] |
| \[\Rightarrow \] \[{{k}^{\frac{1}{x}+\frac{1}{y}}}={{k}^{-\frac{1}{z}}}\] \[[\because {{p}^{m}}\times {{p}^{n}}={{p}^{m+n}}]\] |
| \[\Rightarrow \]\[\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\]\[\Rightarrow \]\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\] |
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