Skip to main content
Lemma
Exponentials and Logarithms Difficulty 4

TMUA practice: Exponentials Logs, problem 4

A TMUA-calibre exponentials logs problem at difficulty 4 of 5, with the full worked solution.

The question
What is the sum of all the solutions of   2+log⁡2(x−2)=log⁡x−28  \;2 + \log_{2}(x - 2) = \log_{x - 2} 8\; in the interval   (2, ∞)  \;(2,\, \infty)\;?
A 358\tfrac{35}{8}
B 55
C 498\tfrac{49}{8}
D 558\tfrac{55}{8}

The correct answer is highlighted. C

Worked solution

Let y=log⁡2(x−2)y = \log_{2}(x - 2). Then   log⁡x−28=log⁡28log⁡2(x−2)=3y  \;\log_{x - 2} 8 = \dfrac{\log_{2} 8}{\log_{2}(x - 2)} = \dfrac{3}{y}\; (change of base).

The equation becomes   2+y=3/y\;2 + y = 3/y, i.e.   y2+2y−3=0\;y^{2} + 2y - 3 = 0, i.e. (y−1)(y+3)=0(y - 1)(y + 3) = 0. So y=1y = 1 or y=−3y = -3.

  • y=1y = 1: log⁡2(x−2)=1⇒x−2=2⇒x=4\log_{2}(x - 2) = 1 \Rightarrow x - 2 = 2 \Rightarrow x = 4.
  • y=−3y = -3: log⁡2(x−2)=−3⇒x−2=1/8⇒x=17/8\log_{2}(x - 2) = -3 \Rightarrow x - 2 = 1/8 \Rightarrow x = 17/8.

Both lie in (2,∞)(2, \infty). (Note 17/8=2.125>217/8 = 2.125 > 2 ✓ and the base x−2=1/8>0x - 2 = 1/8 > 0, ≠1\ne 1 ✓.)

Sum:   4+17/8=32/8+17/8=49/8\;4 + 17/8 = 32/8 + 17/8 = 49/8. This is option C.

The lesson: change-of-base log⁡ab=log⁡cb/log⁡ca\log_{a} b = \log_{c} b / \log_{c} a converts a ‘log-of-a-changing-base’ problem into a polynomial in a single auxiliary variable.