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TMUA 2020 · Paper 1 · Question 15 of 20

TMUA 2020 Paper 1 Question 15

Exponentials and logarithms — Logarithms · a hidden quadratic in log2 x. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 1Exponentials and logarithmsLogarithms · a hidden quadratic in log2 x6 options
Find the positive difference between the two real values of x for which (log2 x)4 + 12(log2(1x))2 − 26 = 0
  1. A4
  2. B16
  3. C154
  4. D174
  5. E25516
  6. F25716
Show the answer and worked solution
answer · C
  1. A4
  2. B16
  3. C154
  4. D174
  5. E25516
  6. F25716
Since log21x = −log2 x, squaring removes the sign. Put u = log2 x: the equation becomes u4 + 12u2 − 64 = 0, which factorises as (u2 + 16)(u2 − 4) = 0. Only u2 = 4 is possible for real u, so u = ± 2 and x = 4 or x = 14. The positive difference is 4 − 14 = 154.