The Chief Examiner reported that this was the most attempted question and majority of the candidates obtained full marks.  Their performance was commended by the Chief Examiner.
In part (a), candidates were reportedly able to simplify 3 to have          which further simplified to 15.

In part (b), majority of the candidates were also reported to have converted both sides of the sign of equality to base 10 and then compared them. That is, 124nfiveconverted to base 10 gave (1 n2) n1)  n0)210.  Simplifying the equation and bringing likes terms together gave n2 + 2n – 63 = 0. Solving this quadratic equation gave n = 7. The other value for n, which was -9, was not acceptable because it was negative.