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Question Number 124705 by ZiYangLee last updated on 05/Dec/20
Given that  ((log a)/(b−c))=((log b)/(c−a))=((log c)/(a−b))  find the value of a^a b^b c^c .
$$\mathrm{Given}\:\mathrm{that} \\ $$$$\frac{\mathrm{log}\:{a}}{{b}−{c}}=\frac{\mathrm{log}\:{b}}{{c}−{a}}=\frac{\mathrm{log}\:{c}}{{a}−{b}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}^{{a}} {b}^{{b}} {c}^{{c}} . \\ $$
Answered by TANMAY PANACEA last updated on 05/Dec/20
p=a^a b^b c^c   logp=aloga+blogb+clogc  =a×k(b−c)+bk(c−a)+ck(a−b)  =0  logp=0=log1  p=1
$${p}={a}^{{a}} {b}^{{b}} {c}^{{c}} \\ $$$${logp}={aloga}+{blogb}+{clogc} \\ $$$$={a}×{k}\left({b}−{c}\right)+{bk}\left({c}−{a}\right)+{ck}\left({a}−{b}\right) \\ $$$$=\mathrm{0} \\ $$$${logp}=\mathrm{0}={log}\mathrm{1} \\ $$$${p}=\mathrm{1} \\ $$
Answered by benjo_mathlover last updated on 05/Dec/20
 { ((log a^(c−a)  = log b^(b−c) )),((log a^(a−b)  = log c^(b−c) )) :}   { (((a^c /a^a ) = (b^b /b^c ) ⇒(ab)^c = a^a b^b )),(((a^a /a^b ) = (c^b /c^c ) ⇒c^c  = (((ac)^b )/a^a ))) :}  then a^a b^b c^c  = (ab)^c .c^c  = (abc)^c
$$\begin{cases}{\mathrm{log}\:{a}^{{c}−{a}} \:=\:\mathrm{log}\:{b}^{{b}−{c}} }\\{\mathrm{log}\:{a}^{{a}−{b}} \:=\:\mathrm{log}\:{c}^{{b}−{c}} }\end{cases} \\ $$$$\begin{cases}{\frac{{a}^{{c}} }{{a}^{{a}} }\:=\:\frac{{b}^{{b}} }{{b}^{{c}} }\:\Rightarrow\left({ab}\right)^{{c}} =\:{a}^{{a}} {b}^{{b}} }\\{\frac{{a}^{{a}} }{{a}^{{b}} }\:=\:\frac{{c}^{{b}} }{{c}^{{c}} }\:\Rightarrow{c}^{{c}} \:=\:\frac{\left({ac}\right)^{{b}} }{{a}^{{a}} }}\end{cases} \\ $$$${then}\:{a}^{{a}} {b}^{{b}} {c}^{{c}} \:=\:\left({ab}\right)^{{c}} .{c}^{{c}} \:=\:\left({abc}\right)^{{c}} \\ $$

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