Question Number 224539 by fantastic last updated on 17/Sep/25
$${a}^{{x}} ={m},\:{a}^{{y}} ={n}\:,{a}^{\mathrm{2}} =\left({m}^{{y}} {n}^{{x}} \right)^{{z}} \\ $$$${prove}\:{xyz}=\mathrm{1} \\ $$
Answered by som(math1967) last updated on 17/Sep/25
$${a}^{{x}} ={m}\Rightarrow{a}^{{xy}} ={m}^{{y}} \\ $$$$\:{a}^{{y}} ={n}\Rightarrow{a}^{{xy}} ={n}^{{x}} \\ $$$$\:{a}^{\mathrm{2}} =\left({m}^{{y}} {n}^{{x}} \right)^{{z}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\left({a}^{{xy}} ×{a}^{{xy}} \right)^{{z}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={a}^{\mathrm{2}{xyz}} \\ $$$$\Rightarrow\mathrm{2}{xyz}=\mathrm{2}\:\:\:\:\therefore{xyz}=\mathrm{1} \\ $$
Commented by fantastic last updated on 17/Sep/25
$${thanks}\:{sir} \\ $$
Answered by Kademi last updated on 17/Sep/25
$$\:\:{a}^{\mathrm{2}} \:=\:\left({m}^{{y}} {n}^{{x}} \right)^{{z}} \\ $$$$\:\:{a}^{\mathrm{2}} \:=\:\left(\left({a}^{{x}} \right)^{{y}} \left({a}^{{y}} \right)^{{x}} \right)^{{z}} \\ $$$$\:\:{a}^{\mathrm{2}} \:=\:\left({a}^{{xy}} {a}^{{xy}} \right)^{{z}} \\ $$$$\:\:{a}^{\mathrm{2}} \:=\:\left({a}^{{xy}+{xy}} \right)^{{z}} \\ $$$$\:\:{a}^{\mathrm{2}} \:=\:\left({a}^{\mathrm{2}{xy}} \right)^{{z}} \\ $$$$\:\:{a}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}{xyz}} \\ $$$$\:\:{a}^{\mathrm{1}} \:=\:{a}^{{xyz}} \\ $$$$\:\:\mathrm{1}\:=\:{xyz}\: \\ $$
Commented by fantastic last updated on 17/Sep/25
$${thanks}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Sep/25
$${a}={m}^{\mathrm{1}/{x}} ,{a}={n}^{\mathrm{1}/{y}} ,{a}=\left({m}^{{y}} {n}^{{x}} \right)^{{z}/\mathrm{2}} \\ $$$${a}={m}^{\frac{\mathrm{1}}{{x}}} ={n}^{\frac{\mathrm{1}}{{y}}} \\ $$$${a}^{{xy}} ={m}^{{y}} ={n}^{{x}} \\ $$$${a}^{\mathrm{2}} =\left({m}^{{y}} {n}^{{x}} \right)^{{z}} \\ $$$${a}^{\mathrm{2}} =\left({a}^{{xy}} .{a}^{{xy}} \right)^{{z}} \\ $$$${a}={a}^{{xyz}} \\ $$$${xyz}=\mathrm{1} \\ $$
Commented by fantastic last updated on 17/Sep/25
$$\nu.\eta\overset{.} {\iota}\subset\epsilon! \\ $$