logxb−c=logyc−a=logza−b=k
Therefore, log10x = k (b -c) ……. (i)
log10y = k (c -a) ……. (ii)
log10z= k (a-b) ……. (iii)
Adding equations (i), (ii) and (iii) we get,
log10x + log10y + log10z = k (b -c) + k (c -a) + k (a-b)
log10xyz = k (b - c + c - a + a - b)
log10xyz = 0
xyz = 100 (changing to exponential form)
xyz = 1