[x=log_(a)(bc),y=log_(b)(ca)" ige "z=log_(c)(ab)" oral chelie "],[" (i) "(1)/(x+1)+(1)/(y+1)+(1)/(z+1)=1quad " (ii) "x+y+z+2=xyz]

If x = log_(a)(bc), y = log_(b)(ca), z = log_(c)(ab) , then find 1/(x+1) + 1/(y+1) + 1/(z+1)

if =log_(a)(bc),y=log_(b)(ca) and z=log_(c)(ab) then (1)/(x+1)+(1)/(y+1)+(1)/(z+1)

If x=log_(a)(bc),y=log_(b)(ca) and z=log_(c )(ab) show that x+y+z+2=xyz

If x=log_(a)(bc),y=log_(b)(ca),z=log_(c)(ab) then (1)/(x+1)+(1)/(y+1)+(1)/(z+1) is equal to

If x=log_(a)bc,y=log_(b)ca,z=log_(c)ab, then the value of (1)/(1+x)+(1)/(1+y)+(1)/(1+z) will be

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)bc = x, log_(b)ca = y and log_(c)ab = z , then the value of (1)/(x +1) + (1)/(y +1) + (1)/(z+1) is equal to

If x=log_(a)(bc),y=log_(b)(ca)" and "z=log_(c)(ab) , then which of the following is equal to 1 ?

If x = log_(a)^(bc) , y = log_(b)^(ca) and z = log_(c)^(ab) then show that frac(1)(x+1)+frac(1)(y+1)+frac(1)(z+1) = 1 , [abc ne 1]