If \( \Large y = \frac{1}{a^{1-log ax}} \) and \( \Large z = \frac{1}{a^{1+log_{a}y}} \) then x is equal to:

From the given relation, we have

\( \Large a = y^{1-log ax} = z^{1-log ay} \)

\( \Large \therefore\ log_{a}a = \left(1-log_{a}x\right)log_{a}y \)

\( \Large and\ log_{a}a = \left(1-log_{a}y\right)log_{a}z \)

=> \( \Large log_{a}y \left(1-log_{a}x\right) = 1 \)

\( \Large and\ log_{a}z \left(1-log_{a}y\right) = 1 \)

=> \( \Large log_{a}y = \frac{1}{1-log_{a}x} \)

\( \Large and\ log_{a}z = \frac{1}{1-log_{a}y} \)

\( \Large \therefore log_{a}z = \frac{1}{1-log_{a}y} = \frac{1}{1-\frac{1}{1-log_{a}x}}\)

= \( \Large \frac{1-log_{a}x}{-log_{a}x} \)

\( \Large Now \frac{1}{1-log_{a}z} = \frac{1}{1 + \frac{1-log_{a}x}{logt_{a}x}}= log_{a}x \)

\( \Large log_{a}x = \frac{1}{1 - log_{a}z} \)

\( \Large x = a \frac{1}{1-log_{a}z} \)