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

$$\mathrm{solve}\mathrm{for}x$$$${\log }_{\mid \sin x\mid }(x^2-8x+23)>\frac{3}{{\log }_2\mid \sin x\mid }$$

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

$$\frac{\log ({\mathrm{x}}^2-8\mathrm{x}+23)}{\log \mid \mathrm{sinx}\mid }>\frac{3\log 2}{\log \mid \mathrm{sinx}\mid }$$$$\log ({\mathrm{x}}^2-8\mathrm{x}+23)>\log 8$$$${\mathrm{x}}^2-8\mathrm{x}+23>8\Rightarrow {\mathrm{x}}^2-8\mathrm{x}+15>0$$$$\mathrm{x}<3\mathrm{or}\mathrm{x}>5$$$$\mathrm{sinx}\ne 0\Rightarrow \mathrm{x}\ne \mathrm{k}\pi ,\mathrm{k}\in \mathbb{Z}$$$$\log \mid \mathrm{sinx}\mid \ne 0\Rightarrow \mid \mathrm{sinx}\mid \ne 1\Rightarrow \mathrm{x}=\frac{(2\mathrm{n}+1)\pi }{2}$$$$\Rightarrow \mathrm{S}=\{\mathrm{x}\in \mathbb{R}-\{\mathrm{k}\pi ,\frac{(2\mathrm{n}+1)\pi }{2}\},\mathrm{k},\mathrm{n}\in \mathbb{Z}\mid \mathrm{x}\in (-\mathrm{\infty },3)\cup (5,\mathrm{\infty })\}$$

Commented by infinityaction last updated on 28/Aug/22

$$checkyoursolution$$

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