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 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

a(b-c) +b(c - a) +c(a-b) is equal to

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=

Simplify: a(b-c)-b(c-a)-c(a-b)

Simplify: a(b-c)+b(c-a)+c(a-b)

If a, b, c are in HP., then a/(b+c),b/(c+a),c/(a+b) are in

If loga=1/2 logb=1/5logc then a^4b^3c^(-2)=

If a,b,c are in G.P then (b-a)/(b-c)+(b+a)/(b+c)=