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Question Number 133936 by liberty last updated on 25/Feb/21

$\log_{x + 8} (x^{2} - 3 x - 4) < 2 . \log_{{(4 - x)}^{2}} (∣ x - 4 ∣)$
	Answered by EDWIN88 last updated on 25/Feb/21
	$log _(x+8) (x^2 −3x−4)<2.log _((4−x)^2 ) (∣x−4∣) { ((x^2 −3x−4>0)),((x+8>0 ; x+8≠1 )),(((4−x)^2 ≠1 , x≠4 )) :} { ((x<−1 ; x>4)),((x>−8)),((x≠4; x≠3; x≠5)) :} ⇔ log _(x+8) (x^2 −3x−4)<2.log _((x−4)^2 ) ∣x−4∣ ⇔log _(x+8) (x^2 −3x−4)<2.(1/2)log _(∣x−4∣) ∣x−4∣ ⇔ log _(x+8) (x^2 −3x−4)<1 ; ((ln (x^2 −3x−4))/(ln (x+8))) <1 ((ln (x^2 −3x−4)−ln (x+8))/(ln (x+8))) <0 numerator :ln (x^2 −3x−4) = ln (x+8) ; x^2 −4x−12=0 x_1 =6 ; x_2 =−2 denumerator : ln (x+8)=0 ; x=−7 solution set is (−8;−7)∪(−2;−1)∪(4;5)∪(5;6)$
	$\log_{x + 8} (x^{2} - 3 x - 4) < 2 . \log_{{(4 - x)}^{2}} (∣ x - 4 ∣)$ ${\begin{cases} x^{2} - 3 x - 4 > 0 \\ x + 8 > 0; x + 8 \neq 1 \\ {(4 - x)}^{2} \neq 1, x \neq 4 \end{cases} {\begin{cases} x < - 1; x > 4 \\ x > - 8 \\ x \neq 4; x \neq 3; x \neq 5 \end{cases}$ $\Leftrightarrow \log_{x + 8} (x^{2} - 3 x - 4) < 2 . \log_{{(x - 4)}^{2}} ∣ x - 4 ∣$ $\Leftrightarrow \log_{x + 8} (x^{2} - 3 x - 4) < 2 . \frac{1}{2} \log_{∣ x - 4 ∣} ∣ x - 4 ∣$ $\Leftrightarrow \log \underset{x + 8}{} (x^{2} - 3 x - 4) < 1; \frac{\ln (x^{2} - 3 x - 4)}{\ln (x + 8)} < 1$ $\frac{\ln (x^{2} - 3 x - 4) - \ln (x + 8)}{\ln (x + 8)} < 0$ $numerator : \ln (x^{2} - 3 x - 4) = \ln (x + 8); x^{2} - 4 x - 12 = 0$ $x_{1} = 6; x_{2} = - 2$ $denumerator : \ln (x + 8) = 0; x = - 7$ $solution set is (- 8; - 7) \cup (- 2; - 1) \cup (4; 5) \cup (5; 6)$
	Answered by bobhans last updated on 25/Feb/21
	$⇔ log _(x+8) (x^2 −3x−4) < 2.log _((4−x)^2 ) (√((4−x)^2 )) log _(x+8) (x^2 −3x−4) < 1 log _(x+8) (x^2 −3x−4) < log _(x+8) (x+8) ⇔ (x+8−1)(x^2 −3x−4−x−8)< 0 (x+7)(x^2 −4x−12) < 0 (x+7)(x−6)(x+2) < 0 (i) x < −7 ∪ −2 < x < 6 (ii) { (((4−x)^2 ≠ 0 ⇒x≠ 4)),(((4−x)^2 ≠ 1 ⇒ x≠ 3; x≠ 5)),((x+8 > 0 ; x > −8)) :} (iii) x^2 −3x−4 > 0 ; (x−4)(x+1)>0 x<−1 ∪ x > 4 solution : (i)∩(ii)∩(iii)$
	$\Leftrightarrow \log_{x + 8} (x^{2} - 3 x - 4) < 2 . \log_{{(4 - x)}^{2}} \sqrt{{(4 - x)}^{2}}$ $\log_{x + 8} (x^{2} - 3 x - 4) < 1$ $\log_{x + 8} (x^{2} - 3 x - 4) < \log_{x + 8} (x + 8)$ $\Leftrightarrow (x + 8 - 1) (x^{2} - 3 x - 4 - x - 8) < 0$ $(x + 7) (x^{2} - 4 x - 12) < 0$ $(x + 7) (x - 6) (x + 2) < 0$ $(i) x < - 7 \cup - 2 < x < 6$ $(ii) {\begin{cases} {(4 - x)}^{2} \neq 0 \Rightarrow x \neq 4 \\ {(4 - x)}^{2} \neq 1 \Rightarrow x \neq 3; x \neq 5 \\ x + 8 > 0; x > - 8 \end{cases}$ $(iii) x^{2} - 3 x - 4 > 0; (x - 4) (x + 1) > 0$ $x < - 1 \cup x > 4$ $solution : (i) \cap (ii) \cap (iii)$
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