If y = (1/(a^(1-loga x))), z = (1/(a^(1-loga y))) and x = a^k, then k = -Maths 9th

If y = (1/(a^(1-loga x))), z = (1/(a^(1-loga y))) and x = a^k, then k = -Maths 9th
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Answer :

(b) \(rac{1}{{1-log_az}}\) y = \(rac{1}{a^{1-log_ax}}\) = \(a^{-(1-log_ax)}\)⇒ logay = \(rac{1}{{1-log_ax}}\) and loga z = \(rac{1}{{1-log_ay}}\)∴ logaz = \(rac{1}{1-\bigg(rac{1}{1-log_ax}\bigg)}\) = \(rac{1-log_ax}{-log_ax}\)⇒ – loga z =  –1 + \(rac{1}{{log_ax}}\) ⇒ \(rac{1}{{log_ax}}\) = 1 - loga z⇒ loga x = \(rac{1}{{1-log_az}}\)⇒ x = \(a^{rac{1}{1-log_az}}\) = ak ⇒ k = \(rac{1}{{1-log_az}}\).
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