Logarithmic Conundrum
Level
pending
Given that a, b, and c form an arithmetic progression and x, y, and z form a geometric progression, what is the sum of (b-c) log x + (c-a) log y + (a-b) log z ?
The answer is 0.
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1 solution
Let a = b+d = c+2d
Let x = u Therefore.. y=ur , x = u r 2
Sub these into the equation:
= -(d)log(x) + (2d)log(y) - (d)log(z) = d(2log(y) - log(x) - log(y)
= d(2log(ur) - log(u) - log( u r 2 )
= d l o g ( u ∗ u r 2 ( u r ) 2 )
= d(0) = 0