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TMUA 2018 · Paper 1 · Question 15 of 20

TMUA 2018 Paper 1 Question 15

Exponentials and logarithms — Exponential equations · hidden quadratic. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 1Exponentials and logarithmsExponential equations · hidden quadratic6 options
Find the sum of the real solutions of the equation: 3x − (√3)x+4 + 20 = 0
  1. A1
  2. B4
  3. C9
  4. Dlog3 20
  5. E2log3 20
  6. F4log3 20
Show the answer and worked solution
answer · E
  1. A1
  2. B4
  3. C9
  4. Dlog3 20
  5. E2log3 20
  6. F4log3 20
Write everything in terms of u = 3x/2. Then 3x = u2 and (√3)x+4 = 3x+42 = 9⋅3x/2 = 9u, so the equation becomes u2 − 9u + 20 = 0, giving u = 4 or u = 5. Both are positive, so both are attainable: x2 = log3 4 or x2 = log3 5. The sum is 2log3 4 + 2log3 5 = 2log3 20.