Math Question and Answers


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 175335 by infinityaction last updated on 27/Aug/22

$solve the inequalities$ $Q . (1) \frac{1 + \log_{a}^{2} x}{1 + \log_{a} x} > 1, 0 < a < 1$ $Q . (2) \log_{x} \frac{4 x + 5}{6 - 5 x} < - 1$
	Answered by blackmamba last updated on 27/Aug/22

	$(1) x > 0$ $l e t \log_{a} x = t$ $\frac{1 + t^{2}}{1 + t} - \frac{1 + t}{1 + t} > 0$ $\frac{t^{2} - t}{t + 1} > 0 \Rightarrow \frac{t (t - 1)}{t + 1} > 0$ $- 1 < t < 0 \lor t > 1$ $- 1 < \log_{a} x < 0 \lor \log_{a} x > 1$ $\log_{a} (\frac{1}{a}) < \log_{a} x < \log_{a} 1 \lor \log_{a} x > \log_{a} a$ $1 < x < \frac{1}{a} \lor 0 < x < a$
	Commented byinfinityaction last updated on 27/Aug/22

	$t h a n k s$
	Answered by mahdipoor last updated on 27/Aug/22
	$2: if a<b ⇒ { ((c^a <c^b if c≥1)),((c^a >c^b if 0<c<1)) :} ((4x+5)/(6−5x))>0⇒x∈(((−5)/4),(6/5)) (iii) ⇒⇒I: 0<x<1 (i) x^∧ (log_x ((4x+5)/(6−5x))) > x^∧ (−1) ⇒((4x+5)/(6−5x))>(1/x) ⇒ 4x^2 +10x−6>0 ⇒x∈R−[−3,(1/2)] (ii) i∩ii∩iii=0.5<x<1 ⇒⇒II: 1≤x (i) x^∧ (log_x ((4x+5)/(6−5x))) ≤ x^∧ (−1) ⇒((4x+5)/(6−5x))≤(1/x) ⇒ 4x^2 +10x−6≤0 ⇒x∈[−3,(1/2)] (ii) i∩ii∩iii=∄ I∩II= 0.5<x<1$
	$2 : i f a < b \Rightarrow {\begin{cases} c^{a} < c^{b} i f c ⩾ 1 \\ c^{a} > c^{b} i f 0 < c < 1 \end{cases}$ $\frac{4 x + 5}{6 - 5 x} > 0 \Rightarrow x \in (\frac{- 5}{4}, \frac{6}{5}) (i i i)$ $\Rightarrow\Rightarrow I : 0 < x < 1 (i)$ $x^{\land} ({l o g}_{x} \frac{4 x + 5}{6 - 5 x}) > x^{\land} (- 1) \Rightarrow \frac{4 x + 5}{6 - 5 x} > \frac{1}{x} \Rightarrow$ $4 x^{2} + 10 x - 6 > 0 \Rightarrow x \in R - [- 3, \frac{1}{2}] (i i)$ $i \cap i i \cap i i i = 0.5 < x < 1$ $\Rightarrow\Rightarrow II : 1 ⩽ x (i)$ $x^{\land} ({l o g}_{x} \frac{4 x + 5}{6 - 5 x}) ⩽ x^{\land} (- 1) \Rightarrow \frac{4 x + 5}{6 - 5 x} ⩽ \frac{1}{x} \Rightarrow$ $4 x^{2} + 10 x - 6 ⩽ 0 \Rightarrow x \in [- 3, \frac{1}{2}] (i i)$ $i \cap i i \cap i i i = ∄$ $I \cap II = 0.5 < x < 1$
	Commented byinfinityaction last updated on 27/Aug/22

	$t h a n k s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum