Question

Sa se determine nr. real x, stiind ca
${2}^{x} - 1$
,
${4}^{x}$
,
${2}^{x + 1} + 3$
sunt trei termeni consecutivi ai unei progresii aritmetice. ​

Answer 1

$\div \: {2}^{x} - 1, {4}^{x} , {2}^{x + 1} + 3 = > {4}^{x} = \frac{ {2}^{x} - 1 + {2}^{x + 1} + 3}{2}$

${2}^{2x} = \frac{ {2}^{x} + {2}^{x + 1} + 2}{2}$

$2 \times {2}^{2x} = {2}^{x} + {2}^{x + 1} + 2$

$2 \times {2}^{2x} - {2}^{x} - {2}^{x + 1} - 2 = 0$

$2 \times { ({2}^{x} )}^{2} - {2}^{x} - {2}^{x} \times 2 - 2 = 0$

${2}^{x} = t \: \: ,t > 0$

$2 {t}^{2} - t - 2t - 2 = 0$

$2 {t}^{2} - 3t - 2 = 0$

$a = 2$

$b = - 3$

$c = - 2$

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

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

$\Delta = 9 + 16 = 25$

$t_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - ( - 3) \pm \sqrt{25} }{2 \times 2} =\frac{3\pm5}{4}$

$t_{1}=\frac{3 + 5}{4} = \frac{8}{4} = 2 > 0 \: = > verifica \: conditia$

$t_{2}=\frac{3 - 5}{4} = - \frac{2}{4} = - \frac{1}{2} < 0 = > nu \: verifica \: conditia$

$= > t = 2 = > {2}^{x} = 2 = > x = 1$