Maths

2^x = 3^y = 12^z

x log2 = y log3 = z log12 = K (where K is a constant)

So, x = K/log2, y = K/log3, z = K/log12

xy = K^2/[(log2)(log3)]

z(x + 2y)

= (K/log12)(K/log2 + 2K/log3)

= (K/log12)[(log3)K + (2log2)K]/[(log2)(log3)]

= (K/log12)[K(log3 + 2log2)]/[(log2)(log3)]

= K^2/[(log2)(log3)]

So xy = z(x + 2y)