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TMUA 2017 · Paper 2 · Question 18 of 20

TMUA 2017 Paper 2 Question 18

Inequalities and reasoning — Taking logarithms to a base less than 1. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 2Inequalities and reasoningTaking logarithms to a base less than 16 options
Consider the following problem:

Solve the inequality (14)n < (132)10, where n is a positive integer.

A student produces the following argument:

(14)n < (132)10
(I)   log12(14)n < log12(132)10
(II)  nlog12(14) < 10log12(132)
(III) n < 10log12(132)log12(14)
(IV)  n < 10×52 = 25
(V)   1 ≤ n ≤ 24

Which step (if any) in the argument is invalid?

  1. AThere are no invalid steps; the argument is correct
  2. BOnly step (I) is invalid; the rest are correct
  3. COnly step (II) is invalid; the rest are correct
  4. DOnly step (III) is invalid; the rest are correct
  5. EOnly step (IV) is invalid; the rest are correct
  6. FOnly step (V) is invalid; the rest are correct
Show the answer and worked solution
answer · B
  1. AThere are no invalid steps; the argument is correct
  2. BOnly step (I) is invalid; the rest are correct
  3. COnly step (II) is invalid; the rest are correct
  4. DOnly step (III) is invalid; the rest are correct
  5. EOnly step (IV) is invalid; the rest are correct
  6. FOnly step (V) is invalid; the rest are correct
The base of the logarithm is 12, which is less than 1, so log12 is a decreasing function and applying it to both sides must reverse the inequality. Step (I) keeps the direction, so that is where it goes wrong. Everything after it is sound: the power rule in (II), dividing by log1214 = 2 (a positive number) in (III), the arithmetic in (IV), and reading off the integers in (V). Writing the original as 2−2n < 2−50 confirms the true answer is n > 25, not n < 25.