So → a=log3(k) (that is, the logarithm of k based on three) b=log4(k)c=log6(k).

so→ 1/a = logk(3) 1/b = logk(4) 1/c = logk(6)

so→2/c = logk(36)2/a = logk(9) 1/b = logk(4)

So →2/c=2/a+ 1/b

(Make trouble with computers ...)

If you change the seventh question, you can slowly calculate (1) so that 3 x = t (2) and 2x = t.