Question

If $y=a^{\left(\frac{1}{1-{\log }_{a}x}\right)}$ and $z=a^{\left(\frac{1}{1-{\log }_{a}y}\right)}$, then relation between $x$ and $z$ is

• A. $x=a^{\left(\frac{1}{1-{\log }_{a}z}\right)}$
• B. $x=a^{\left(\frac{1}{1+{\log }_{a}z}\right)}$
• C. $x=a\left(\frac{1}{1-{\log }_{a}z}\right)$
• D. $x=a\left(\frac{1}{1+{\log }_{a}z}\right)$

Answer 1

The correct option is A $x=a^{\left(\frac{1}{1-{\log }_{a}z}\right)}$

$y=a^{\frac{1}{1-{\log }_{a}x}}\Rightarrow {\log }_{a}y=\frac{1}{1-{\log }_{a}x}$, taking log both sides on base 'a'

$z=a^{\frac{1}{1-{\log }_{a}y}}\Rightarrow {\log }_{a}z=\frac{1}{1-{\log }_{a}y}=\frac{1}{1-\frac{1}{1-{\log }_{a}x}}$

$\Rightarrow {\log }_{a}z=\frac{1-{\log }_{a}x}{1-{\log }_{a}x-1}\Rightarrow -{\log }_{a}x{\log }_{a}z=1-{\log }_{a}x$

$\Rightarrow {\log }_{a}x=\frac{1}{1-{\log }_{a}z}\Rightarrow x=a^{\frac{1}{1-{\log }_{a}z}}$