[tex] 1)log_{x + 1}( {x}^{2} + 3x - 14) = 2[/tex]
Condiții de existență :
[tex]x + 1 > 0[/tex]
[tex]x + 1 \neq1[/tex]
[tex] {x}^{2} + 3x - 14 > 0[/tex]
[tex] {(x + 1)}^{2} = {x}^{2} + 3x - 14[/tex]
[tex] {x}^{2} + 2x + 1 = {x}^{2} + 3x - 14[/tex]
[tex] {x}^{2} - {x}^{2} + 2x - 3x + 1 + 14 = 0[/tex]
[tex] - x + 15 = 0[/tex]
[tex] - x = - 15 \: | \times ( - 1)[/tex]
[tex]x = 15 \: verifica \: conditiile[/tex]
[tex]S=\left\{15\right\}[/tex]
[tex]2) log_{4}( {3}^{x} - 1) = log_{4}( - {3}^{x} + 14) [/tex]
Condiții de existență :
[tex] {3}^{x} - 1 > 0[/tex]
[tex] - {3}^{x} + 14 > 0[/tex]
[tex] {3}^{x} - 1 = - {3}^{x} + 14[/tex]
[tex] {3}^{x} + {3}^{x} = 14 + 1[/tex]
[tex]2 \times {3}^{x} = 15[/tex]
[tex] {3}^{x} = \frac{15}{2} [/tex]
[tex]ln {3}^{x} = ln \frac{15}{2} [/tex]
[tex]xln3 = ln \frac{15}{2} [/tex]
[tex]x = \frac{ln \frac{15}{2} }{ln3} \:\:vetifica \:conditiile[/tex]
[tex]S=\left\{\frac{ln \frac{15}{2} }{ln3}\right\}[/tex]
[tex]3) log_{3}( {x}^{2} - 14x + 22) = 2[/tex]
Condiția de existență :
[tex] {x}^{2} - 14x + 22 > 0[/tex]
[tex] {x}^{2} - 14x + 22 = {3}^{2} [/tex]
[tex] {x}^{2} - 14x + 22 = 9[/tex]
[tex] {x}^{2} - 14x + 22 - 9 = 0[/tex]
[tex] {x}^{2} - 14x + 13 = 0[/tex]
[tex]a = 1[/tex]
[tex]b = - 14[/tex]
[tex]c = 13[/tex]
[tex]\Delta = {b}^{2} - 4ac[/tex]
[tex]\Delta = {( - 14)}^{2} - 4 \times 1 \times 13[/tex]
[tex]\Delta = 196 - 52[/tex]
[tex]\Delta = 144[/tex]
[tex]x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 14) \pm \sqrt{144} }{2 \times 1} = \frac{14 \pm12}{2} [/tex]
[tex]x_{1} = \frac{14 + 12}{2} = \frac{26}{2} = 13 \: \: verifica \: conditia[/tex]
[tex]x_{2} = \frac{14 - 12}{2} = \frac{2}{2} = 1 \: verifica \: conditia[/tex]
[tex]S=\left\{1,13\right\}[/tex]
[tex]4) log_{x}(2 {x}^{2} - 3x) = 1[/tex]
Condiții de existență :
[tex]x > 0[/tex]
[tex]x \: \neq \: 1[/tex]
[tex]2 {x}^{2} - 3x > 0[/tex]
[tex]2 {x}^{2} - 3x = {x}^{1} [/tex]
[tex]2 {x}^{2} - 3x = x[/tex]
[tex]2 {x}^{2} - 3x - x = 0[/tex]
[tex]2 {x}^{2} - 4x = 0 \: | \div 2[/tex]
[tex] {x}^{2} - 2x = 0[/tex]
[tex]x(x - 2) = 0[/tex]
[tex]x_{1} = 0 \: nu \: verifica \: conditiile[/tex]
[tex]x - 2 = 0 = > x_{2} = 2 \: verifica \: conditiile[/tex]
[tex]S=\left\{2\right\}[/tex]
