Question Number 124705 by ZiYangLee last updated on 05/Dec/20
$$\mathrm{Given}\:\mathrm{that} \\ $$$$\frac{\mathrm{log}\:{a}}{{b}−{c}}=\frac{\mathrm{log}\:{b}}{{c}−{a}}=\frac{\mathrm{log}\:{c}}{{a}−{b}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}^{{a}} {b}^{{b}} {c}^{{c}} . \\ $$
Answered by TANMAY PANACEA last updated on 05/Dec/20
$${p}={a}^{{a}} {b}^{{b}} {c}^{{c}} \\ $$$${logp}={aloga}+{blogb}+{clogc} \\ $$$$={a}×{k}\left({b}−{c}\right)+{bk}\left({c}−{a}\right)+{ck}\left({a}−{b}\right) \\ $$$$=\mathrm{0} \\ $$$${logp}=\mathrm{0}={log}\mathrm{1} \\ $$$${p}=\mathrm{1} \\ $$
Answered by benjo_mathlover last updated on 05/Dec/20
$$\begin{cases}{\mathrm{log}\:{a}^{{c}−{a}} \:=\:\mathrm{log}\:{b}^{{b}−{c}} }\\{\mathrm{log}\:{a}^{{a}−{b}} \:=\:\mathrm{log}\:{c}^{{b}−{c}} }\end{cases} \\ $$$$\begin{cases}{\frac{{a}^{{c}} }{{a}^{{a}} }\:=\:\frac{{b}^{{b}} }{{b}^{{c}} }\:\Rightarrow\left({ab}\right)^{{c}} =\:{a}^{{a}} {b}^{{b}} }\\{\frac{{a}^{{a}} }{{a}^{{b}} }\:=\:\frac{{c}^{{b}} }{{c}^{{c}} }\:\Rightarrow{c}^{{c}} \:=\:\frac{\left({ac}\right)^{{b}} }{{a}^{{a}} }}\end{cases} \\ $$$${then}\:{a}^{{a}} {b}^{{b}} {c}^{{c}} \:=\:\left({ab}\right)^{{c}} .{c}^{{c}} \:=\:\left({abc}\right)^{{c}} \\ $$