If `loga/(b-c) = logb/(c-a) = logc/(a-b)`, then `a^(b+c).b^(c+a).c^(a+b)`=

If (loga)/(b-c) = (logb)/(c-a) = (logc)/(a-b) , then prove that a^(a)b^(b)c^(c)=1 .

if loga/(b-c)=logb/(c-a)=logc/(a-b) then find the value of a^ab^bc^c

Let (log a)/( b-c) = (logb)/(c-a) = (log c)/(a-b) STATEMENT 1: a^(a) b^(b) c^(c) = 1 STATEMENT 2: a^(b+c) b^(c +a) c^(a + b) = 1

If (a(b+c-a))/(loga) = (b(c+a-b))/(logb) = (c(a+b-c))/(logc) , then prove that b^(c ) . c^(b) = a^(c ).c^(a) = a^(b).b^(a) .

if x =log_a (bc), y= log_b (ca) and z=log_c(ab) then 1/(x+1)+1/(y+1)+1/(z+1)

If log_a x, log_b x, log_c x are in A.P then c^2=