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TMUA 2016 · Paper 1 · Question 16 of 20

TMUA 2016 Paper 1 Question 16

Exponentials and logarithms — Simultaneous logarithmic equations. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 1Exponentials and logarithmsSimultaneous logarithmic equations5 options
Given the simultaneous equations log10 2 + log10(y−1) = 2log10 x log10(y + 3 − 3x) = 0 the values of y are
  1. A52 ± 3√52
  2. B3 ± √3
  3. C7 ± 3√3
  4. D3,  9
  5. E1,  13
Show the answer and worked solution
answer · C
  1. A52 ± 3√52
  2. B3 ± √3
  3. C7 ± 3√3
  4. D3,  9
  5. E1,  13
Strip the logs. The first equation is log10(2(y−1)) = log10 x2, so 2(y−1) = x2. The second says the argument equals 100 = 1, so y + 3 − 3x = 1, that is y = 3x − 2. Substituting, 2(3x − 3) = x2 gives x2 − 6x + 6 = 0 and x = 3 ± √3, both positive so both are legitimate. Then y = 3x − 2 = 7 ± 3√3, and both make y − 1 > 0, so neither is rejected.