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Question Number 33100 by artibunja last updated on 10/Apr/18

	Answered by MJS last updated on 10/Apr/18

	$48 x^{2} - 8 x - 8 > 0 \Rightarrow$ $\Rightarrow x < - \frac{1}{3} \lor x > \frac{1}{2}$ $42 x^{2} + 34 x - 8 > 0 \Rightarrow$ $\Rightarrow x < - 1 \lor x > \frac{4}{21}$ $\Rightarrow x < - 1 \lor x > \frac{1}{2}$ $\frac{\ln (48 x^{2} - 8 x - 8)}{\ln \frac{3}{4}} ⩽ \frac{\ln (42 x^{2} + 34 x - 8)}{\ln \frac{3}{4}}$ $\ln (48 x^{2} - 8 x - 8) ⩾ \ln (42 x^{2} + 34 x - 8)$ $48 x^{2} - 8 x - 8 ⩾ 42 x^{2} + 34 x - 8$ $6 x^{2} - 42 x ⩾ 0$ $(x - 7) x ⩾ 0$ $x ⩾ 0 \land (x - 7) ⩾ 0 \Rightarrow x ⩾ 7$ $x ⩽ 0 \land (x - 7) ⩽ 0 \Rightarrow x ⩽ 0$ $answer is$ $x < - 1 \lor x ⩾ 7$
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