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  4. If 4x=16y=64z , then

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Q. If $4^x=16^y=64^z$ , then

Sequences and Series

Solution:

$4^{x}= 16^{y} = 64^{z} = K$
$ \Rightarrow 4= K^{\frac{1}{x}}: 16 = K^{\frac{1}{y}}, 64= K^{\frac{1}{z}} $
$\Rightarrow 4\times 64 = K^{\frac{1}{x}} : K^{\frac{1}{z}} = K^{\frac{2}{y}} = K^{\frac{1}{x}+\frac{1}{z}} $
$ \Rightarrow 256 = K^{\frac{1}{x}} \cdot K^{\frac{1}{z}} = K^{\frac{2}{y}} = K^{\frac{1}{x}+\frac{1}{z}} $
$ \Rightarrow \frac{2}{y} = \frac{1}{x} +\frac{1}{z}$
$\Rightarrow \frac{1}{x}\cdot\frac{1}{y}\cdot\frac{1}{z}$ are in $A.P$.
$ \left(\because256=16^{2} = K^{\frac{2}{y}}\right)$