Question

If `loga/(b-c) = logb/(c-a) = logc/(a-b)`, then `a^(b+c).b^(c+a).c^(a+b)`=
A. `a^(b)b^(c)c^(a)=1`
B. `a^(a)b^(b)c^(c)=1`
C. `root(a)(a) root(b)(b) root(c)(c) = 1`
D. none of these

Answer 1

Correct Answer - B
We have,
`("log"a)/(b-c) = ("log"b)/(c-a) = ("log"c)/(a-b) = lambda("say")`
`rArr a = 10^(lambda (b-c)), b = 10^(lambda(c-a)), c = 10^(lambda(a-b))`
`rArr a^(a)b^(b)c^(c) = 10^(lambdaa(b-c)+lambda b(c-a) +lambdac(a-b))=10^(0) =1`