solve-for-x-log-sinx-x-2-8x-23-gt-3-log-2-sinx-

Question Number 175369 by infinityaction last updated on 28/Aug/22

solve for x

\log_{∣ \sin x ∣} (x^{2} - 8 x + 23) > \frac{3}{\log_{2} ∣ \sin x ∣}

Answered by floor(10²Eta[1]) last updated on 28/Aug/22

\frac{\log (x^{2} - 8 x + 23)}{\log ∣ sinx ∣} > \frac{3 \log 2}{\log ∣ sinx ∣}

\log (x^{2} - 8 x + 23) > \log 8

x^{2} - 8 x + 23 > 8 \Rightarrow x^{2} - 8 x + 15 > 0

x < 3 or x > 5

sinx \neq 0 \Rightarrow x \neq k π, k \in Z

\log ∣ sinx ∣\neq 0 \Rightarrow∣ sinx ∣\neq 1 \Rightarrow x = \frac{(2 n + 1) π}{2}

\Rightarrow S = {x \in R - {k π, \frac{(2 n + 1) π}{2}}, k, n \in Z ∣ x \in (- \infty, 3) \cup (5, \infty)}

Commented by infinityaction last updated on 28/Aug/22

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