If y=a11-logaxand z=a11-logay, then x is equal to

∵y=a11−logax⇒logay=11−logax...........(i)and z=a11−logayor logaz=11−logay..........(2)From Eqs.(i) and (ii), we getlogaz=11−(11−logax)=1−1logax⇒1logax=(1−logaz)⇒logax=1(1−logaz)∴x=a1a1−logaz

If y=a11-logaxand z=a11-logay, then x is equal to

∵y=a11−logax⇒logay=11−logax...........(i)and z=a11−logayor logaz=11−logay..........(2)From Eqs.(i) and (ii), we getlogaz=11−(11−logax)=1−1logax⇒1logax=(1−logaz)⇒logax=1(1−logaz)∴x=a1a1−logaz

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If $y = a^{\frac{1}{1 - \log_{a} x}}$ and $z = a^{\frac{1}{1 - \log_{a} y}}$ , then x is equal to

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$a^{\frac{1}{1 + \log_{a} z}}$

$a^{\frac{1}{2 + \log_{a} z}}$

$a^{\frac{1}{1 - \log_{a} z}}$

$a^{\frac{1}{2 - \log_{a} z}}$

answer is C.

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Detailed Solution

$∵ y = a^{\frac{1}{1 - \log_{a} x}}$
$\Rightarrow \log_{a} y = \frac{1}{1 - \log_{a} x} ........... (i)$
and $z = a^{\frac{1}{1 - \log_{a} y}}$
or $\log_{a} z = \frac{1}{1 - \log_{a} y} .......... (2)$
From Eqs.(i) and (ii), we get
$\log_{a} z = \frac{1}{1 - (\frac{1}{1 - \log_{a} x})} = 1 - \frac{1}{\log_{a} x}$
$\Rightarrow \frac{1}{\log_{a} x} = (1 - \log_{a} z) \Rightarrow \log_{a} x = \frac{1}{(1 - \log_{a} z)}$

$∴ x = a^{\frac{1}{a^{1 - \log_{a} z}}}$