a , b , c are positive reals such that
b − c lo g a = c − a lo g b = a − b lo g c .
What is the value of
a a × b b × c c ?
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Nice solution!
let a^a=k, b^b=k, c^c=k.. apply log principals and we goo
Why do you take all the three to be "k"? Could you elaborate?
You must instead take the equal expressions to be "K"
Downvoted!!
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a loga / a(b-c) = b logb / b(c-a) = c*logc / c(a-b) = k (some value)
Sum of numerators/ Sum of denominators = k {If a/x= b/y= c/z = k, then a= kx, b= ky, c= kz Therefore, k= (a+b+c)/(x+y+z)}
So, we can say that a log a+b log b+ c log c= k (a(b-c) + b(c-a) + c(a-b)) a log a+b log b+ c log c= k (0) a log a+b log b+ c log c= 0 log a^a.b^b.c^c= 0
a^a.b^b.c^c= 1