Candidates found the (a) part of the question very simple to solve. They were able to rearrange the log to get x = 10/x-3 » x = 5 or – 2 since x ≠ -ve, x = 5.
The (b) part on binomial expansion however was poorly attempted by most candidates. They needed to understand that to find the value of (1.94)7,
2 – 3x = 1.94 = 2 – 0.06 where 3x = 0.06 \ x = 0.02 should be substituted for x in the expansion of (2 – 3x)7
= 128 – 1344x + 6048x2 – 15120x3 to get 103.4182
@ 103.418 to three decimal places.
