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TMUA 2016 · Paper 1 · Question 11 of 20

TMUA 2016 Paper 1 Question 11

Exponentials and logarithms — Disguised quadratic in 22x · change of base. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 1Exponentials and logarithmsDisguised quadratic in 22x · change of base6 options
The real roots of the equation 42x + 12 = 22x+3 are p and q, where p > q.

The value of p − q can be expressed as

  1. A34
  2. B1
  3. C4
  4. D−12 + log1032
  5. Elog10 3log10 4
  6. Flog10 3log10 2
Show the answer and worked solution
answer · E
  1. A34
  2. B1
  3. C4
  4. D−12 + log1032
  5. Elog10 3log10 4
  6. Flog10 3log10 2
Everything is a power of 2: 42x = 24x = (22x)2 and 22x+3 = 8⋅ 22x. With u = 22x the equation is u2 − 8u + 12 = 0, so u = 2 or u = 6. Then 22x = 2 gives x = 12, and 22x = 6 gives x = 12log2 6, which is the larger root. So p − q = 12(log2 6 − 1) = 12log2 3 = log10 32log10 2 = log10 3log10 4.