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

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(B) दिया गया है: $y = \frac{1}{a^{1-\log _{a} x}} \implies \log _{a} y = \frac{1}{1-\log _{a} x}$.साथ ही,$z = \frac{1}{a^{1-\log _{a} y}} \implies \lo

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

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(B) दिया गया है: $y = \frac{1}{a^{1-\log _{a} x}} \implies \log _{a} y = \frac{1}{1-\log _{a} x}$.साथ ही,$z = \frac{1}{a^{1-\log _{a} y}} \implies \log _{a} z = \frac{1}{1-\log _{a} y}$.$\log _{a} y$ का मान $\log _{a} z$ के समीकरण में रखने पर:$\log _{a} z = \frac{1}{1 - \frac{1}{1-\log _{a} x}} = \frac{1}{\frac{1-\log _{a} x - 1}{1-\log _{a} x}} = \frac{1-\log _{a} x}{-\log _{a} x}$.समीकरण को पुनर्व्यवस्थित करने पर:$-\log _{a} z = \frac{1-\log _{a} x}{\log _{a} x} = \frac{1}{\log _{a} x} - 1$.अतः,$\frac{1}{\log _{a} x} = 1 - \log _{a} z$.इसका अर्थ है कि $\log _{a} x = \frac{1}{1 - \log _{a} z}$.चूंकि $x = a^{k}$ दिया गया है,इसलिए $\log _{a} x = k$.अतः,$k = \frac{1}{1 - \log _{a} z}$.

यदि $y = \frac{1}{a^{1-\log _{a} x}}$,$z = \frac{1}{a^{1-\log _{a} y}}$ और $x = a^{k}$ है,तो $k =$

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