If y = frac{1}{a^{1-log _{a} x}},z = frac{1}{ | vedclass

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(B) Given: $y = \frac{1}{a^{1-\log _{a} x}} \implies \log _{a} y = \frac{1}{1-\log _{a} x}$.Also,$z = \frac{1}{a^{1-\log _{a} y}} \implies \log _{a} z

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$\frac{1}{1-\log _{a} z}$

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(B) Given: $y = \frac{1}{a^{1-\log _{a} x}} \implies \log _{a} y = \frac{1}{1-\log _{a} x}$.Also,$z = \frac{1}{a^{1-\log _{a} y}} \implies \log _{a} z = \frac{1}{1-\log _{a} y}$.Substitute $\log _{a} y$ into the expression for $\log _{a} z$:$\log _{a} z = \frac{1}{1 - \frac{1}{1-\log _{a} x}} = \frac{1}{\frac{1-\log _{a} x - 1}{1-\log _{a} x}} = \frac{1-\log _{a} x}{-\log _{a} x}$.Rearranging the equation:$-\log _{a} z = \frac{1-\log _{a} x}{\log _{a} x} = \frac{1}{\log _{a} x} - 1$.Therefore,$\frac{1}{\log _{a} x} = 1 - \log _{a} z$.This implies $\log _{a} x = \frac{1}{1 - \log _{a} z}$.Since $x = a^{k}$,we have $\log _{a} x = k$.Thus,$k = \frac{1}{1 - \log _{a} z}$.

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

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