If logxb-c=logyc-a=logza-b, then xyz=?
0
1
-1
2
Explanation for the correct option:
Given, logxb-c=logyc-a=logza-b
Let this be logxb-c=logyc-a=logza-b=k
So, logx=k(b-c),logy=k(c-a) and logz=k(a-b)
Taking antilog on both sides, we get,
x=ek(a-b),y=ek(b-c) and z=ek(c-a) [∵logba=c⇒a=bc]
Substituting these values in xyz, we get
ek(a-b)·ekb-c.ekc-a=ek(a-b+b-c+c-a)[∵am×an=am+n]=ek(0)=e0=1
Hence, option(B) i.e. 1, is the correct answer.