Solve-log-2x-3-x-2-lt-1-

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

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!