Question

Exercitiul 18 subpunctele c,d,e

Answer 1

$c) {4}^{x} + 4 = 5 \times {2}^{x}$

${( {2}^{x}) }^{2} - 5 \times {2}^{x} + 4 = 0$

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

${t}^{2} - 5t +4 = 0$

${t}^{2} - 4t - t + 4 = 0$

$t(t - 4) - (t - 4) = 0$

$(t - 4)(t - 1) = 0$

$t - 4 = 0 = > t_{1} = 4 = > {2}^{x} = 4 = >x_{1} = 2$

$t - 1 = 0 = > t_{2}= 1 = > {2}^{x} = 1 = > x_{2} = 0$

$S=\left\{0,2\right\}$

$d) {2}^{x} + {4}^{x} = 72$

${2}^{x} + {( {2}^{x}) }^{2} = 72$

${ ({2}^{x} )}^{2} + {2}^{x} - 72 = 0$

${t}^{2} + t - 72 = 0$

$a = 1$

$b = 1$

$c = - 72$

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

$\Delta = {1}^{2} - 4 \times 1 \times ( - 72)$

$\Delta = 1 + 288$

$\Delta = 289$

$t_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a} = \frac{ - 1 \pm \sqrt{289} }{2 \times 1} = \frac{ - 1 \pm17}{2}$

$t_{1} = \frac{ - 1 + 17}{2} = \frac{16}{2} = 8 = > {2}^{x} = 8 = > x_{1}= 3$

$t_{2} = \frac{ - 1 - 17}{2} = - \frac{18}{2} = - 9 < 0 \: nu \: convine$

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

$e) {9}^{x} + 9 = 10 \times {3}^{x}$

${ ({3}^{x} )}^{2} - 10 \times {3}^{x} + 9 = 0$

${3}^{x} = t \: \: \: ,t > 0$

${t}^{2} - 10t + 9 = 0$

${t}^{2} - 9t - t + 9 = 0$

$t(t - 9) - (t - 9) = 0$

$(t - 9)(t - 1) = 0$

$t - 9 = 0 = > t_{1} = 9 = > {3}^{x} = 9 = > x_{1} = 2$

$t - 1 = 0 = > t_{2}= 1 = > {3}^{x} = 1 = > x_{2} = 0$

$S=\left\{0,2\right\}$