Question 1046135
{{{(loga)/(y-z)=(logb)/(z-x)=(logc)/(x-y) }}}

<===>{{{log(a^(1/(y-z)))= log(b^(1/(z-x))) = log(c^(1/(x-y))) }}}

<===>{{{a^(1/(y-z))= b^(1/(z-x)) = c^(1/(x-y)) }}}

===> {{{a^(z-x) = b^(y-z)}}},  {{{b^(x-y) = c^(z-x)}}}, and {{{a^(x-y) = c^(y-z)}}}.

Multiplying the corresponding sides of the 1st and 3rd equations give

{{{a^(z-y) = (bc)^(y-z)}}},

<===> {{{1/a^(y-z) = (bc)^(y-z)}}}  ===>   {{{1 = (abc)^(y-z)}}}.

Since {{{y - z <>0 }}} for all real numbers y and z, there is only one possibility for the value of abc, and that is, {{{highlight (abc = 1)}}}.


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