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If y = \(\frac{1}{a^{1-log_ax}}\), z = \(\frac{1}{a^{1-log_ay}}\) and x = ak, then k =

(a) \(\frac{1}{a^{1-log_az}}\)

(b) \(\frac{1}{{1-log_az}}\)

(c) \(\frac{1}{{1+log_za}}\)

(d) \(\frac{1}{{1-log_za}}\)

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(b) \(\frac{1}{{1-log_az}}\)

 y = \(\frac{1}{a^{1-log_ax}}\) = \(a^{-(1-log_ax)}\)

⇒ logay = \(\frac{1}{{1-log_ax}}\) and loga z = \(\frac{1}{{1-log_ay}}\)

∴ logaz = \(\frac{1}{1-\bigg(\frac{1}{1-log_ax}\bigg)}\) = \(\frac{1-log_ax}{-log_ax}\)

⇒ – loga z =  –1 + \(\frac{1}{{log_ax}}\) ⇒ \(\frac{1}{{log_ax}}\) = 1 - loga z

⇒ loga x = \(\frac{1}{{1-log_az}}\)

⇒ x = \(a^{\frac{1}{1-log_az}}\) = ak ⇒ k = \(\frac{1}{{1-log_az}}\).

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