Question

Rezolvați in Mulțimea numerelor reale ecuațiile :
2lgx=lg(4x-3)

Answer 1

$2lgx = lg(4x - 3)$

Condiții de existență :

$x>0$

$4x-3>0$

$lg {x}^{2} = lg(4x - 3)$

${x}^{2} = 4x - 3$

${x}^{2} - 4x + 3 = 0$

$M1) {x}^{2} - 4x + 3 = 0$

${x}^{2} - 3x - x + 3 = 0$

$x(x - 3) - (x - 3) = 0$

$(x - 3)(x - 1) = 0$

$x - 3 = 0 = > x_{1} = 3\:verifica \: conditiile$

$x - 1 = 0 = > x_{2}= 1\:verifica\:conditiile$

$S=\left\{1,3\right\}$

$M2) {x}^{2} - 4x + 3 = 0$

$a = 1$

$b = - 4$

$c = 3$

$\Delta = {b}^{2} - 4ac$

$\Delta = {( - 4)}^{2} - 4 \times 1 \times 3$

$\Delta = 16 - 12$

$\Delta = 4$

$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 4) \pm \sqrt{4} }{2 \times 1} = \frac{4 \pm2}{2}$

$x_{1} = \frac{4 + 2}{2} = \frac{6}{2} = 3\:verifica\:conditiile$

$x_{2} = \frac{4 - 2}{2} = \frac{2}{2} = 1\:verifica\:conditiile$

$S=\left\{1,3\right\}$