Q13/087) Prove that if x = log_a(bc),y = log_b(ca),z = log_c(ab) then the value of xyz - x - y - z is 2?

we get

bc = a^x ..1
ca = b^y...2
ab = c ^z ...3

from (3)

c^(xyz) = (ab)^(xy)
= a^(xy) * b^(xy)
= (a^x)^y) * (b^y)^x)
= (bc)^y ( ac)^x ( from 1 and 2)
= c^y c^x ( b^y)(a^x)
= c^y c^x (ac) (bc)
= c^y c^x c^2(ab)
= c^ x c^y c^2 c^z
= c^(x+y+z+2)

hence xyz = x + y + z + 2 or
xyz - x - y - z = 2

Alternatively we can solve as below

\

multiply (1) by a

abc = a^(x+1) => a = (abc)^(1/(x+1) ..4

similarly

b = (abc)^(1/(y+1) ..5

c = (abc)^(1/(z+1) .. 6

multiply to get

abc = (abc)^(1/(x+1) + 1/(y+1) + 1(z+1))

or (1/(x+1) + 1/(y+1) + 1(z+1)) = 1

or (y+1)(z+1) + (x+1)(z+1) + (x+1)(y+1) = ( x+1)(y+1) (z+1)

or yz + y + z + 1 + xz + z + x + 1 + xy + x + y + 1 = xyz + xy + yz + xz + x + y+ z + 1

or x+y + z + 2 = xyz