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Question Number 42549 by Tawa1 last updated on 27/Aug/18

$$\mathrm{If}\mathrm{y}={\mathrm{a}}^{\frac{1}{1-{\log }_{\mathrm{a}}\mathrm{x}}}\mathrm{and}\mathrm{z}={\mathrm{a}}^{\frac{1}{1-{\log }_{\mathrm{a}}\mathrm{y}}}.\mathrm{show}\mathrm{that}\mathrm{x}={\mathrm{a}}^{\frac{1}{1-{\log }_{\mathrm{a}}\mathrm{z}}}$$

Answered by math1967 last updated on 27/Aug/18

$${log}_{a}y=\frac{1}{1-{log}_{a}x}\Rightarrow {log}_{a}x=1-\frac{1}{{log}_{a}y}$$$$similarly{log}_{a}y=1-\frac{1}{{log}_{a}z}$$$$\therefore {log}_{a}x=1-\frac{1}{1-\frac{1}{{log}_{a}z}}=\frac{-1}{{log}_{a}z-1}=\frac{1}{1-{log}_{a}z}$$$$\therefore x=a^{\frac{1}{1-{log}_{a}z}}$$

Commented by Tawa1 last updated on 27/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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