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TMUA 2020 · Paper 1 · Question 7 of 20

TMUA 2020 Paper 1 Question 7

Exponentials and logarithms — Indices · simultaneous equations in powers of 2. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 1Exponentials and logarithmsIndices · simultaneous equations in powers of 26 options
Given that 23x = 8(y+3) and 4(x+1) = 16(y+1)8(y+3) what is the value of x + y?
  1. A−23
  2. B−22
  3. C−15
  4. D−14
  5. E−11
  6. F−10
Show the answer and worked solution
answer · A
  1. A−23
  2. B−22
  3. C−15
  4. D−14
  5. E−11
  6. F−10
Write everything as a power of 2. The first equation gives 23x = 23(y+3), so 3x = 3y + 9 and x = y + 3. The second gives 22(x+1) = 24(y+1) − 3(y+3) = 2y−5, so 2x + 2 = y − 5. Substituting x = y+3 gives 2y + 8 = y − 5, so y = −13 and x = −10. Hence x + y = −23.