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Question Number 206332 by MATHEMATICSAM last updated on 12/Apr/24
If log(a + b + c) = loga + logb + logc then prove that log(((2a)/(1 − a^2 )) + ((2b)/(1 − b^2 )) + ((2c)/(1 − c^2 ))) = log(((2a)/(1 − a^2 ))) + log(((2b)/(1 − b^2 ))) + log(((2c)/(1 − c^2 ))).
$$\mathrm{If}\:\mathrm{log}\left({a}\:+\:{b}\:+\:{c}\right)\:=\:\mathrm{log}{a}\:+\:\mathrm{log}{b}\:+\:\mathrm{log}{c}\: \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right)\:=\: \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right). \\ $$
Answered by TheHoneyCat last updated on 12/Apr/24
Exponantiating we get the equivalent problem: Show that (a+b+c)=abc ⇒((2a)/(1 − a^2 )) + ((2b)/(1 − b^2 )) + ((2c)/(1 − c^2 )) = ((2a)/(1 − a^2 ))×((2b)/(1 − b^2 ))×((2c)/(1 − c^2 )) ⇔(2/(1−a^2 ))(a+ ((b−a^2 b)/(1 − b^2 )) + ((c−a^2 c)/(1 − c^2 )) ) = ((2a)/(1 − a^2 ))×((2b)/(1 − b^2 ))×((2c)/(1 − c^2 )) ⇔a+ ((b−a^2 b)/(1 − b^2 )) + ((c−a^2 c)/(1 − c^2 )) = a((2b)/(1 − b^2 ))×((2c)/(1 − c^2 )) ∨a^2 =1 ⇔a−ab^2 +b−a^2 b + ((c−a^2 c−b^2 c+a^2 b^2 c)/(1 − c^2 )) =2ab×((2c)/(1 − c^2 )) ∨a=1∨b=1 ⇔a−ab^2 −ac^2 +ab^2 c^2 +b−a^2 b −bc^2 +a^2 bc^2 +c−a^2 c−b^2 c+a^2 b^2 c =2^2 abc ∨a=1∨b=1∨c=1 ⇔a+b+c −(ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c)+ abc(bc+ac+ab)=2^2 abc∨a=1∨b=1∨c=1 ⇔ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c =abc(ab+ac+bc−3) ∨a=1∨b=1∨c=1 ⇔ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c =a^2 b+a^2 c+abc+ab^2 +abc+b^2 c+abc+ac^2 +bc^2 −3abc ∨a=1∨b=1∨c=1 ⇔ab^2 +ac^2 +a^2 b+bc^2 +a^2 c+b^2 c =a^2 b+a^2 c+ab^2 +b^2 c+ac^2 +bc^2 ∨a=1∨b=1∨c=1 ⇔True ∨a=1∨b=1∨c=1 ⇔True So reading things backward... If log(a + b + c) = loga + logb + logc log(((2a)/(1 − a^2 )) + ((2b)/(1 − b^2 )) + ((2c)/(1 − c^2 ))) = log(((2a)/(1 − a^2 ))) + log(((2b)/(1 − b^2 ))) + log(((2c)/(1 − c^2 ))) _□
$$\mathrm{Exponantiating}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{equivalent} \\ $$$$\mathrm{problem}: \\ $$$${Show}\:{that}\:\left({a}+{b}+{c}\right)={abc}\: \\ $$$$\Rightarrow\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }×\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }×\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\frac{\mathrm{2}}{\mathrm{1}−{a}^{\mathrm{2}} }\left({a}+\:\frac{{b}−{a}^{\mathrm{2}} {b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\:+\:\frac{{c}−{a}^{\mathrm{2}} {c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }×\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }×\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} } \\ $$$$\Leftrightarrow{a}+\:\frac{{b}−{a}^{\mathrm{2}} {b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\:+\:\frac{{c}−{a}^{\mathrm{2}} {c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\:=\:{a}\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }×\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\:\vee{a}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Leftrightarrow{a}−{ab}^{\mathrm{2}} +{b}−{a}^{\mathrm{2}} {b}\:+\:\frac{{c}−{a}^{\mathrm{2}} {c}−{b}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\mathrm{2}{ab}×\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\:\vee{a}=\mathrm{1}\vee{b}=\mathrm{1} \\ $$$$\Leftrightarrow{a}−{ab}^{\mathrm{2}} −{ac}^{\mathrm{2}} +{ab}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}−{a}^{\mathrm{2}} {b}\:−{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {bc}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+{c}−{a}^{\mathrm{2}} {c}−{b}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}\:=\mathrm{2}^{\mathrm{2}} {abc}\: \\ $$$$\:\:\:\:\:\vee{a}=\mathrm{1}\vee{b}=\mathrm{1}\vee{c}=\mathrm{1} \\ $$$$\Leftrightarrow{a}+{b}+{c}\:−\left({ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c}\right)+ \\ $$$$\:\:\:\:\:{abc}\left({bc}+{ac}+{ab}\right)=\mathrm{2}^{\mathrm{2}} {abc}\vee{a}=\mathrm{1}\vee{b}=\mathrm{1}\vee{c}=\mathrm{1} \\ $$$$\Leftrightarrow{ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c} \\ $$$$\:\:\:\:\:={abc}\left({ab}+{ac}+{bc}−\mathrm{3}\right)\:\vee{a}=\mathrm{1}\vee{b}=\mathrm{1}\vee{c}=\mathrm{1} \\ $$$$\Leftrightarrow{ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c} \\ $$$$\:\:\:\:\:={a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{abc}+{ab}^{\mathrm{2}} +{abc}+{b}^{\mathrm{2}} {c}+{abc}+{ac}^{\mathrm{2}} +{bc}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:−\mathrm{3}{abc}\:\vee{a}=\mathrm{1}\vee{b}=\mathrm{1}\vee{c}=\mathrm{1} \\ $$$$\Leftrightarrow{ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}+{bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c} \\ $$$$\:\:\:\:\:={a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{ac}^{\mathrm{2}} +{bc}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\vee{a}=\mathrm{1}\vee{b}=\mathrm{1}\vee{c}=\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{True}\:\vee{a}=\mathrm{1}\vee{b}=\mathrm{1}\vee{c}=\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{True} \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{reading}\:\mathrm{things}\:\mathrm{backward}… \\ $$$$\mathrm{If}\:\mathrm{log}\left({a}\:+\:{b}\:+\:{c}\right)\:=\:\mathrm{log}{a}\:+\:\mathrm{log}{b}\:+\:\mathrm{log}{c}\: \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right)\:=\: \\ $$$$\mathrm{log}\left(\frac{\mathrm{2}{a}}{\mathrm{1}\:−\:{a}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{b}}{\mathrm{1}\:−\:{b}^{\mathrm{2}} }\right)\:+\:\mathrm{log}\left(\frac{\mathrm{2}{c}}{\mathrm{1}\:−\:{c}^{\mathrm{2}} }\right)\:\:\:_{\Box} \\ $$

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