[tex] \div \: {2}^{x} - 1, {4}^{x} , {2}^{x + 1} + 3 = > {4}^{x} = \frac{ {2}^{x} - 1 + {2}^{x + 1} + 3}{2} [/tex]
[tex] {2}^{2x} = \frac{ {2}^{x} + {2}^{x + 1} + 2}{2} [/tex]
[tex] 2 \times {2}^{2x} = {2}^{x} + {2}^{x + 1} + 2[/tex]
[tex]2 \times {2}^{2x} - {2}^{x} - {2}^{x + 1} - 2 = 0[/tex]
[tex]2 \times { ({2}^{x} )}^{2} - {2}^{x} - {2}^{x} \times 2 - 2 = 0[/tex]
[tex] {2}^{x} = t \: \: ,t > 0[/tex]
[tex]2 {t}^{2} - t - 2t - 2 = 0[/tex]
[tex]2 {t}^{2} - 3t - 2 = 0[/tex]
[tex]a = 2[/tex]
[tex]b = - 3[/tex]
[tex]c = - 2[/tex]
[tex]\Delta = {b}^{2} - 4ac[/tex]
[tex]\Delta = {( - 3)}^{2} - 4 \times 2 \times ( - 2)[/tex]
[tex]\Delta = 9 + 16 = 25[/tex]
[tex]t_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 3) \pm \sqrt{25} }{2 \times 2} =\frac{3\pm5}{4}[/tex]
[tex]t_{1}=\frac{3 + 5}{4} = \frac{8}{4} = 2 > 0 \: = > verifica \: conditia[/tex]
[tex]t_{2}=\frac{3 - 5}{4} = - \frac{2}{4} = - \frac{1}{2} < 0 = > nu \: verifica \: conditia[/tex]
[tex] = > t = 2 = > {2}^{x} = 2 = > x = 1[/tex]
