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TMUA 2016 · Paper 2 · Question 7 of 20

TMUA 2016 Paper 2 Question 7

Exponentials and logarithms — Logarithm laws · assembling the steps of a proof. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 2Exponentials and logarithmsLogarithm laws · assembling the steps of a proof8 options
The four real numbers a, b, c, and d are all greater than 1.

Suppose that they satisfy the equation logc d = (loga b)2.

Use some of the lines given to construct a proof that, in this case, it follows that (×)   logb d = (loga b)(loga c)

(1) Let x = loga b and y = loga c

(2)  d = (cx)2
(3)  d = c(x2)
(4)  d = bxy

(5)  d = (ay)(x2)
(6)  d = ((ay)x)2
(7)  d = (ax)xy

(8)  d = a(y2 x)
(9)  d = a(x2 y)

  1. A(1). Then (2), so (6), so (8), so (7), and therefore (4), hence (*) as required.
  2. B(1). Then (2), so (7), so (8), so (6), and therefore (4), hence (*) as required.
  3. C(1). Then (3), so (5), so (9), so (7), and therefore (4), hence (*) as required.
  4. D(1). Then (3), so (7), so (9), so (5), and therefore (4), hence (*) as required.
  5. E(1). Then (4), so (5), so (9), so (7), and therefore (3), hence (*) as required.
  6. F(1). Then (4), so (6), so (8), so (7), and therefore (2), hence (*) as required.
  7. G(1). Then (4), so (7), so (8), so (6), and therefore (2), hence (*) as required.
  8. H(1). Then (4), so (7), so (9), so (5), and therefore (3), hence (*) as required.
Show the answer and worked solution
answer · C
  1. A(1). Then (2), so (6), so (8), so (7), and therefore (4), hence (*) as required.
  2. B(1). Then (2), so (7), so (8), so (6), and therefore (4), hence (*) as required.
  3. C(1). Then (3), so (5), so (9), so (7), and therefore (4), hence (*) as required.
  4. D(1). Then (3), so (7), so (9), so (5), and therefore (4), hence (*) as required.
  5. E(1). Then (4), so (5), so (9), so (7), and therefore (3), hence (*) as required.
  6. F(1). Then (4), so (6), so (8), so (7), and therefore (2), hence (*) as required.
  7. G(1). Then (4), so (7), so (8), so (6), and therefore (2), hence (*) as required.
  8. H(1). Then (4), so (7), so (9), so (5), and therefore (3), hence (*) as required.
Follow the chain forwards from what you are given. With x = loga b and y = loga c, the hypothesis logc d = x2 says d = c(x2), which is line (3). Since c = ay, that becomes d = (ay)(x2), line (5), and so d = a(x2 y), line (9). Regrouping the index as x⋅ xy gives d = (ax)xy, line (7), and ax = b, so d = bxy, line (4). Taking logb of that gives logb d = xy, which is (*). The order is (1), (3), (5), (9), (7), (4).