Unable to solve logarithm question Given (a(b+c-a))/(log a)=(b(c+a-b))/(log b)=(c(a+b-c))/(log c)

Sariah Mcguire

Sariah Mcguire

Answered question

2022-10-30

Unable to solve logarithm question
Given
a ( b + c a ) log a = b ( c + a b ) log b = c ( a + b c ) log c
To prove:
a b b a = b c c b = c a a c
To prove:
a b b a = b c c b = c a a c
What i tried is
log ( a z ) = a ( b + c a )
and similarly for other two. I am unable to break this down further! please help!

Answer & Explanation

Annabella Ferguson

Annabella Ferguson

Beginner2022-10-31Added 15 answers

log a = λ a ( b + c a )
and so on give:
b log a + a log b = λ ( a b ( b + c a ) + a b ( a + c b ) ) = 2 λ a b c
that is symmetric in a , b , c, so by exponentiating the previous line
a b b a = a c c a = b c c b
follows.

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