Question

$lg(2x+6)-lg(2x-3) ^{2} =1$

Answer 1

Condiţii de existenţă: $\left \{ {{2x+6\ \textgreater \ 0} \atop {(2x-3)^2 \ \textgreater \ 0}} \right. \Leftrightarrow \left \{ {{2x\ \textgreater \ -6} \atop {2x-3 \neq 0}} \right. \Leftrightarrow \left \{ {{x\ \textgreater \ -3} \atop {x \neq \frac{3}{2} }} \right.$

Folosim formula $\log_ax-\log_ay=\log_a\frac{x}{y}.$.
[tex]\lg(2x+6)-\lg(2x-3)^2=1 \\ \\ \Rightarrow \lg \frac{2x+6}{(2x-3)^2}=1 \\ \\ \Rightarrow \frac{2x-6}{(2x-3)^2} =10 \\ \\ \Rightarrow 2x+6 = 10(2x-3)^2 \\ \\ \Rightarrow x+3 = 5(2x-3)^2 \\ \\ \Rightarrow x+3 = 5(4x^2-12x+9) \\ \\ \Rightarrow x+3 = 20x^2-60x+45\\ \\ \Rightarrow 20x^2-60x-x+45-3 = 0\\ \\ \Rightarrow 20x^2-61x+42 = 0\\ \\ \Delta = 61 \cdot 61 - 4 \cdot 20 \cdot 42 \\ \\ \Delta = 361 \Rightarrow \sqrt{\Delta} = 19. \\ \\[/tex]
$\left \{ {{x_1 = \frac{61+19}{40}} \atop {x_2 = \frac{61-19}{40}} \right. \Rightarrow \left \{ {{x_1 = 2} \atop {x_2 = \frac{21}{20}} \right.$