Equations & Inequations

Algebra Level 3

log ⁡ 1 3 ( x 2 + x + 1 ) + 1 > 0 \large \log_\frac 13 \left(x^2+x+1\right) + 1 > 0

Find the set of solutions to the inequality above.

( − 2 , 1 ) (-2, 1) R \mathbb R ( − ∞ , − 2 ) ∪ ( 1 , ∞ ) (-\infty, - 2) \cup (1, \infty) [ − 1 , 2 ] [-1, 2]

Aakhyat Singh by Aakhyat Singh , 20, India

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1 solution

log ⁡ 1 3 ( x 2 + x + 1 ) + 1 > 0 log ⁡ 1 3 ( x 2 + x + 1 ) > − 1 x 2 + x + 1 > ( 1 3 ) − 1 = 3 x 2 + x − 2 > 0 ( x + 2 ) ( x − 1 ) > 0 x ∈ ( − 2 , 1 ) \begin{aligned} \log_\frac 13 \left(x^2+x+1\right) + 1 & > 0 \\ \log_\frac 13 \left(x^2+x+1\right) & > - 1 \\ x^2+x+1 & > \left(\frac 13\right)^{-1} = 3 \\ x^2 + x - 2 & > 0 \\ (x+2)(x-1) & > 0 \\ x & \in \boxed{(-2, 1)} \end{aligned}

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In the last 3 inequalities, it should be < rather than >. Otherwise, a nice solution!

Judy Gu - 3 years, 8 months ago

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