MCQ
If ${a^{1/x}} = {b^{1/y}} = {c^{1/z}}$ and ${b^2} = ac$ then $x + z = $
- A$y$
- ✓$2y$
- C$2xyz$
- DNone of these
${b^2} = ac \Rightarrow {({k^y})^2} = {k^x}.{k^z}$ $ \Rightarrow $ ${k^{2y}} = {k^{x + z}} \Rightarrow x + z = 2y$.
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