If {a^{1/x}} = {b^{1/y}} = {c^{1/z}} and {b^2} = ac then x + z =

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Basic of Logarithms

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If ${a^{1/x}} = {b^{1/y}} = {c^{1/z}}$ and ${b^2} = ac$ then $x + z = $

A

$y$

B

$2y$

C

$2xyz$

D

None of these

Solution

(b) ${a^{1/x}} = {b^{1/y}} = {c^{1/z}} = k\,({\rm{say}}) \Rightarrow a = {k^x},\,b = {k^y},\,c = {k^z}$

${b^2} = ac \Rightarrow {({k^y})^2} = {k^x}.{k^z}$ $ \Rightarrow $ ${k^{2y}} = {k^{x + z}} \Rightarrow x + z = 2y$.

Standard 11

Mathematics

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