Solve : log_(2x+3) x^2


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Question Number 21357 by Tinkutara last updated on 21/Sep/17

$Solve : \log_{2 x + 3} x^{2} < 1$
	Answered by dioph last updated on 21/Sep/17
	$2x + 3 > 0 ⇒ x > −(3/2) 2x + 3 ≠ 1 ⇒ x ≠ −1 ((log x^2 )/(log 2x+3)) < 1 Case 1: x < −1 (2x + 3 < 1) log 2x + 3 < 0 ⇒ log x^2 > log 2x+3 ⇒ log (x^2 /(2x+3)) > 0 ⇒ (x^2 /(2x+3)) > 1 ⇒ x^2 −2x − 3 > 0 ⇒ x ∈ (−∞,−1) ∪ (3, +∞) but x < −1 ⇒ x ∈ (−∞,−1) Case 2: x > −1 (2x+3 > 1) log 2x+3 > 0 ⇒ log x^2 < log 2x+3 ⇒ log (x^2 /(2x+3)) < 0 ⇒ (x^2 /(2x+3)) < 1 ⇒ x^2 −2x−3 < 0 ⇒ x ∈ (−1,3) x > −1 ⇒ x ∈ (−1, 3) putting together: x ∈ (−(3/2), 3) − {−1}$
	$2 x + 3 > 0 \Rightarrow x > - \frac{3}{2}$ $2 x + 3 \neq 1 \Rightarrow x \neq - 1$ $\frac{\log x^{2}}{\log 2 x + 3} < 1$ $Case 1 : x < - 1 (2 x + 3 < 1)$ $\log 2 x + 3 < 0$ $\Rightarrow \log x^{2} > \log 2 x + 3$ $\Rightarrow \log \frac{x^{2}}{2 x + 3} > 0$ $\Rightarrow \frac{x^{2}}{2 x + 3} > 1$ $\Rightarrow x^{2} - 2 x - 3 > 0$ $\Rightarrow x \in (- \infty, - 1) \cup (3, + \infty)$ $but x < - 1 \Rightarrow x \in (- \infty, - 1)$ $Case 2 : x > - 1 (2 x + 3 > 1)$ $\log 2 x + 3 > 0$ $\Rightarrow \log x^{2} < \log 2 x + 3$ $\Rightarrow \log \frac{x^{2}}{2 x + 3} < 0$ $\Rightarrow \frac{x^{2}}{2 x + 3} < 1$ $\Rightarrow x^{2} - 2 x - 3 < 0$ $\Rightarrow x \in (- 1, 3)$ $x > - 1 \Rightarrow x \in (- 1, 3)$ $putting together :$ $x \in (- \frac{3}{2}, 3) - {- 1}$
	Commented by Tinkutara last updated on 21/Sep/17

	$Thank you very much Sir!$
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