SOLUTION: Kindly help me solving it i tried my best but I couldn't If {{{(loga)/(y-z)=(logb)/(z-x)=(logc)/(x-y) }}}the value of abc is? Kindly provide me the solution steps thereof. Itried

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Kindly help me solving it i tried my best but I couldn't If {{{(loga)/(y-z)=(logb)/(z-x)=(logc)/(x-y) }}}the value of abc is? Kindly provide me the solution steps thereof. Itried      Log On


   

Question 1046135: Kindly help me solving it i tried my best but I couldn't
If %28loga%29%2F%28y-z%29=%28logb%29%2F%28z-x%29=%28logc%29%2F%28x-y%29+the value of abc is? Kindly provide me the solution steps thereof.
Itried it in the following manner
If loga=logb=logc
then obviously a=b=c
So abc=aaa
Or abc=3a
But that is not the correct answer.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
%28loga%29%2F%28y-z%29=%28logb%29%2F%28z-x%29=%28logc%29%2F%28x-y%29+
<===>
<===>a%5E%281%2F%28y-z%29%29=+b%5E%281%2F%28z-x%29%29+=+c%5E%281%2F%28x-y%29%29+
===> a%5E%28z-x%29+=+b%5E%28y-z%29, b%5E%28x-y%29+=+c%5E%28z-x%29, and a%5E%28x-y%29+=+c%5E%28y-z%29.
Multiplying the corresponding sides of the 1st and 3rd equations give
a%5E%28z-y%29+=+%28bc%29%5E%28y-z%29,
<===> 1%2Fa%5E%28y-z%29+=+%28bc%29%5E%28y-z%29 ===> 1+=+%28abc%29%5E%28y-z%29.
Since y+-+z+%3C%3E0+ for all real numbers y and z, there is only one possibility for the value of abc, and that is, highlight+%28abc+=+1%29.

(You can also do the similar thing with the corresponding sides of the 2nd and the 3rd equations, and arrive at the same conclusion.)