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Question Number 173687 by mnjuly1970 last updated on 16/Jul/22

Q : ((ln(a ))/(c−b)) =((ln(b))/(a−c)) = ((ln(c))/(b−a)) ⇒ a^( a) . b^( b) . c^( c) =?

$$ \\ $$$$\:\:\:{Q}\::\:\:\:\:\:\frac{{ln}\left({a}\:\right)}{{c}−{b}}\:=\frac{{ln}\left({b}\right)}{{a}−{c}}\:=\:\frac{{ln}\left({c}\right)}{{b}−{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:{a}^{\:{a}} .\:{b}^{\:{b}} .\:{c}^{\:{c}} \:=? \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 17/Jul/22

((ln(a ))/(c−b)) =((ln(b))/(a−c)) = ((ln(c))/(b−a)) ((ln(a ))/(c−b)) =((ln(b))/(a−c)) (a−c) ln a = (c−b) ln b ln a^(a−c) =ln b^(c−b) a^(a−c) = b^(c−b) (a^a /a^c )=(b^c /b^b ) a^a b^b =(ab)^c ...................(i) ((ln(b))/(a−c)) = ((ln(c))/(b−a)) (b−a) ln b = (a−c) ln c ln b^(b−a) =ln c^(a−c) b^(b−a) = c^(a−c) (b^b /b^a ) = (c^a /c^c ) b^b c^c =(bc)^a .................(ii) ((ln(a ))/(c−b)) = ((ln(c))/(b−a)) (b−a) ln a = (c−b) ln c ln a^(b−a) =ln c^(c−b) a^(b−a) = c^(c−b) (a^b /a^a )=(c^c /c^b ) a^a c^c =(ac)^b .................(iii) (i)×(ii)×(iii): (a^a b^b )(b^b c^c )(a^a c^c )=(ab)^c (bc)^a (ac)^b (a^a b^b c^c )^2 =a^(b+c) b^(a+c) c^(a+b) a^a b^b c^c =a^((b+c)/2) b^((a+c)/2) c^((a+b)/2) Not complete...

$$\:\:\:\:\:\frac{{ln}\left({a}\:\right)}{{c}−{b}}\:=\frac{{ln}\left({b}\right)}{{a}−{c}}\:=\:\frac{{ln}\left({c}\right)}{{b}−{a}} \\ $$$$\:\:\frac{{ln}\left({a}\:\right)}{{c}−{b}}\:=\frac{{ln}\left({b}\right)}{{a}−{c}} \\ $$$$\left({a}−{c}\right)\:\mathrm{ln}\:{a}\:=\:\left({c}−{b}\right)\:\mathrm{ln}\:{b} \\ $$$$\mathrm{ln}\:{a}^{{a}−{c}} =\mathrm{ln}\:{b}^{{c}−{b}} \\ $$$${a}^{{a}−{c}} =\:{b}^{{c}−{b}} \\ $$$$\frac{{a}^{{a}} }{{a}^{{c}} }=\frac{{b}^{{c}} }{{b}^{{b}} } \\ $$$${a}^{{a}} {b}^{{b}} =\left({ab}\right)^{{c}} ...................\left({i}\right) \\ $$$$\frac{{ln}\left({b}\right)}{{a}−{c}}\:=\:\frac{{ln}\left({c}\right)}{{b}−{a}} \\ $$$$\left({b}−{a}\right)\:\mathrm{ln}\:{b}\:=\:\left({a}−{c}\right)\:\mathrm{ln}\:{c} \\ $$$$\mathrm{ln}\:{b}^{{b}−{a}} =\mathrm{ln}\:{c}^{{a}−{c}} \:\:\:\: \\ $$$$\:{b}^{{b}−{a}} =\:\:{c}^{{a}−{c}} \:\:\:\: \\ $$$$\frac{{b}^{{b}} }{{b}^{{a}} }\:=\:\frac{{c}^{{a}} }{{c}^{{c}} } \\ $$$${b}^{{b}} {c}^{{c}} =\left({bc}\right)^{{a}} .................\left({ii}\right) \\ $$$$\:\:\frac{{ln}\left({a}\:\right)}{{c}−{b}}\:=\:\frac{{ln}\left({c}\right)}{{b}−{a}} \\ $$$$\left({b}−{a}\right)\:\mathrm{ln}\:{a}\:=\:\left({c}−{b}\right)\:\mathrm{ln}\:{c} \\ $$$$\mathrm{ln}\:{a}^{{b}−{a}} =\mathrm{ln}\:{c}^{{c}−{b}} \\ $$$$\:\:{a}^{{b}−{a}} =\:{c}^{{c}−{b}} \\ $$$$\frac{{a}^{{b}} }{{a}^{{a}} }=\frac{{c}^{{c}} }{{c}^{{b}} } \\ $$$${a}^{{a}} {c}^{{c}} =\left({ac}\right)^{{b}} .................\left({iii}\right) \\ $$$$\left({i}\right)×\left({ii}\right)×\left({iii}\right): \\ $$$$\left({a}^{{a}} {b}^{{b}} \right)\left({b}^{{b}} {c}^{{c}} \right)\left({a}^{{a}} {c}^{{c}} \right)=\left({ab}\right)^{{c}} \left({bc}\right)^{{a}} \left({ac}\right)^{{b}} \\ $$$$\left({a}^{{a}} {b}^{{b}} {c}^{{c}} \right)^{\mathrm{2}} ={a}^{{b}+{c}} {b}^{{a}+{c}} {c}^{{a}+{b}} \\ $$$${a}^{{a}} {b}^{{b}} {c}^{{c}} ={a}^{\frac{{b}+{c}}{\mathrm{2}}} {b}^{\frac{{a}+{c}}{\mathrm{2}}} {c}^{\frac{{a}+{b}}{\mathrm{2}}} \\ $$$$\mathcal{N}{ot}\:{complete}... \\ $$

Answered by som(math1967) last updated on 16/Jul/22

((ln(a))/(c−b))=((ln(b))/(a−c))=((ln(c))/(b−a))=k (let) ln(a)=k(c−b) ⇒a=e^(k(c−b)) ⇒a^a =e^(k(ac−ab)) ln(b)=k(a−c)⇒b^b =e^(k(ba−bc)) ln(c)=k(b−a)⇒c^c =e^(k(cb−ca)) ∴a^a .b^b .c^c =e^(k(ac−ab)+k(ba−bc)+k(cb−ca)) =e^0 =1

$$\:\frac{{ln}\left({a}\right)}{{c}−{b}}=\frac{{ln}\left({b}\right)}{{a}−{c}}=\frac{{ln}\left({c}\right)}{{b}−{a}}={k}\:\left({let}\right) \\ $$$${ln}\left({a}\right)={k}\left({c}−{b}\right) \\ $$$$\Rightarrow{a}={e}^{{k}\left({c}−{b}\right)} \Rightarrow{a}^{{a}} ={e}^{{k}\left({ac}−{ab}\right)} \\ $$$${ln}\left({b}\right)={k}\left({a}−{c}\right)\Rightarrow{b}^{{b}} ={e}^{{k}\left({ba}−{bc}\right)} \\ $$$${ln}\left({c}\right)={k}\left({b}−{a}\right)\Rightarrow{c}^{{c}} ={e}^{{k}\left({cb}−{ca}\right)} \\ $$$$\therefore{a}^{{a}} .{b}^{{b}} .{c}^{{c}} \\ $$$$={e}^{{k}\left({ac}−{ab}\right)+{k}\left({ba}−{bc}\right)+{k}\left({cb}−{ca}\right)} \\ $$$$={e}^{\mathrm{0}} =\mathrm{1} \\ $$

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