Question

If $y=a^{\frac{1}{1-{\log }_{a}x}}$ and $z=a^{\frac{1}{1-{\log }_{a}y}}$, then prove that $x=a^{\frac{1}{1-{\log }_{a}z}}$ .

Answer 1

${\log }_{a}y=\frac{1}{1-{\log }_{a}x}$, therefore, $1-{\log }_{a}y=1-\frac{1}{1-{\log }_{a}x}=\frac{-{\log }_{a}x}{1-{\log }_{a}x}$
or $\frac{1}{1-{\log }_{a}y}=\frac{1-{\log }_{a}x}{-{\log }_{a}x}$
But $z=a^{\frac{1}{1-{\log }_{a}y}}$
$\Rightarrow {\log }_{a}z=\frac{1}{1-{\log }_{a}y}=-\frac{1}{{\log }_{a}x}+1$
$\Rightarrow \frac{1}{{\log }_{a}x}=1-{\log }_{a}z$
$\Rightarrow {\log }_{a}x=\frac{1}{1-{\log }_{a}z}$
$\Rightarrow x=a^{\frac{1}{1-{\log }_{a}z}}$