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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-

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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 ))).
Iflog(a+b+c)=loga+logb+logcthenprovethatlog(2a1−a2+2b1−b2+2c1−c2)=log(2a1−a2)+log(2b1−b2)+log(2c1−c2).
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 ))) _□
Exponantiatingwegettheequivalentproblem:Showthat(a+b+c)=abc⇒2a1−a2+2b1−b2+2c1−c2=2a1−a2×2b1−b2×2c1−c2⇔21−a2(a+b−a2b1−b2+c−a2c1−c2)=2a1−a2×2b1−b2×2c1−c2⇔a+b−a2b1−b2+c−a2c1−c2=a2b1−b2×2c1−c2∨a2=1⇔a−ab2+b−a2b+c−a2c−b2c+a2b2c1−c2=2ab×2c1−c2∨a=1∨b=1⇔a−ab2−ac2+ab2c2+b−a2b−bc2+a2bc2+c−a2c−b2c+a2b2c=22abc∨a=1∨b=1∨c=1⇔a+b+c−(ab2+ac2+a2b+bc2+a2c+b2c)+abc(bc+ac+ab)=22abc∨a=1∨b=1∨c=1⇔ab2+ac2+a2b+bc2+a2c+b2c=abc(ab+ac+bc−3)∨a=1∨b=1∨c=1⇔ab2+ac2+a2b+bc2+a2c+b2c=a2b+a2c+abc+ab2+abc+b2c+abc+ac2+bc2−3abc∨a=1∨b=1∨c=1⇔ab2+ac2+a2b+bc2+a2c+b2c=a2b+a2c+ab2+b2c+ac2+bc2∨a=1∨b=1∨c=1⇔True∨a=1∨b=1∨c=1⇔TrueSoreadingthingsbackward…Iflog(a+b+c)=loga+logb+logclog(2a1−a2+2b1−b2+2c1−c2)=log(2a1−a2)+log(2b1−b2)+log(2c1−c2)◻

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