Question Number 42549 by Tawa1 last updated on 27/Aug/18
$$\mathrm{If}\:\:\:\mathrm{y}\:=\:\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{1}\:−\:\mathrm{log}_{\mathrm{a}} \mathrm{x}}} \:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\:\:\mathrm{z}\:=\:\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{1}\:−\:\mathrm{log}_{\mathrm{a}} \mathrm{y}}} \:\:.\:\:\mathrm{show}\:\mathrm{that}\:\:\:\:\:\mathrm{x}\:=\:\mathrm{a}^{\frac{\mathrm{1}}{\mathrm{1}\:−\:\mathrm{log}_{\mathrm{a}} \mathrm{z}}} \\ $$
Answered by math1967 last updated on 27/Aug/18
$${log}_{{a}} {y}=\frac{\mathrm{1}}{\mathrm{1}−{log}_{{a}} {x}}\Rightarrow{log}_{{a}} {x}=\mathrm{1}−\frac{\mathrm{1}}{{log}_{{a}} {y}} \\ $$$${similarly}\:\:{log}_{{a}} {y}=\mathrm{1}−\frac{\mathrm{1}}{{log}_{{a}} {z}} \\ $$$$\therefore{log}_{{a}} {x}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{log}_{{a}} {z}}}=\frac{−\mathrm{1}}{{log}_{{a}} {z}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{1}−{log}_{{a}} {z}} \\ $$$$\therefore{x}={a}^{\frac{\mathrm{1}}{\mathrm{1}−{log}_{{a}} {z}}} \\ $$
Commented by Tawa1 last updated on 27/Aug/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$