Maths Olympiad Prep

Track / Stage 3 / 226 of 260 #226 of 1964

Problem 226

AMC 10/12, early questions
Algebra Difficulty 3.7 Multiple choice China Mathematical Competition · China

Suppose log⁡a(2x2+x−1)>log⁡a2−1\log_a (2x^2 + x - 1) > \log_a 2 - 1. Then the range of xx is:

Pick one

Official solution

From
{x>0,x≠1,2x2+x−1>0 \begin{cases} x > 0, \\ x \ne 1, \\ 2x^2 + x - 1 > 0 \end{cases}
we get x>12x > \frac{1}{2}, x≠1x \ne 1.

Furthermore,
log⁡a(2x2+x−1)>log⁡a2−1⇒log⁡a(2x3+x2−x)>log⁡a2⇒{0<x<1,2x3+x2−x<2,\log_a (2x^2 + x - 1) > \log_a 2 - 1 \Rightarrow \log_a (2x^3 + x^2 - x) > \log_a 2 \Rightarrow \begin{cases} 0 < x < 1, \\ 2x^3 + x^2 - x < 2, \end{cases} or {x>1,2x3+x2−x>2.\begin{cases} x > 1, \\ 2x^3 + x^2 - x > 2. \end{cases} Then we have

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