If logxb-c=logyc-a=logza-b, then the value of xb+c·yc+a·za+b is:
1
2
0
-1
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(b-c),y=ek(c-a) and z=ek(a-b) [∵logba=c⇒a=bc]
Substituting these values in xb+c·yc+a·za+b, we get,
ek(b-c)b+c·ek(c-a)c+a·ek(a-b)a+b=ek(b2-a2)·ek(c2-a2)·ek(a2-b2)[∵(am)n=amn]=ek(b2-a2+c2-a2+a2-b2)[∵am·an=am+n]=ek(0)=1
Hence, option(A) i.e. 1 is the correct answer.