Question

Решите показательные уравнения

Answer 1

$1) {2}^{ {x}^{2} - 5x + 6 } = 1 \\ {x}^{2} - 5x + 6 = 0 \\ d = 25 - 24 = 1 \\ x1 = (5 + 1) \div 2 = 3 \\ x2 = 2$

ответ: 2,3.

$2) {(0.4)}^{x - 1} = {(6.25)}^{6x - 5 } \\ \\ {( \frac{2}{5}) }^{x - 1} = {( \frac{25}{4}) }^{6x - 5} \\ {( \frac{2}{5} )}^{x - 1} = {( \frac{2}{5}) }^{ - 2(6x - 5)} \\ x - 1 = - 12x + 10 \\ 13x = 11 \\ x = \frac{11}{13}$

$3) {5}^{2x - 1} ( {5}^{2} - 3) = 550 \\ {5}^{2x - 1} \times 22 = 550 \\ {5}^{2x - 1} = 25 \\ 2x - 1 = 2 \\ x = 1.5$

$4)2 \times {3}^{x - 1} - {3}^{x - 2} = {5}^{x - 2} + 4 \times {5}^{x - 3} \\ {3}^{x - 2} (2 \times 3 - 1) = {5}^{x - 3} (5 + 4) \\ {3}^{x - 2} \times 5 = {5}^{x - 3} \times 9 \\ \frac{ {3}^{x - 2} }{ {5}^{x - 3} } = \frac{9}{5} \\ \frac{ {3}^{x - 2} }{ {5}^{x - 3} } = \frac{3 \times 3}{5} \\ \frac{ {3}^{x - 2} }{ {5}^{x - 3} \times 3 } = \frac{3}{5} \\ \frac{ {3}^{x - 3} }{ {5}^{x - 3} } = \frac{3}{5} \\ x - 3 = 1 \\ x = 4$

$5)4 \times {2}^{2x} + 3 \times {2}^{x} - 1 =$

Замена:

${2}^{x} = t \\ 4 {t}^{2} + 3t - 1 = 0 \\ d = 9 + 16 = 25 \\ t1 = ( - 3 + 5) \div 8 = \frac{1}{4} \\ t2 = - 1$

${2}^{x} = {2}^{ - 2} \\ x = - 2 \\ {2}^{x} = - 1$

нет корней.

ответ:-2.

$6) {8}^{x} - {4}^{x + 0.5} - {2}^{x} + 2 = 0$

Замена:

${2}^{x} = t \\ {t}^{3} - 2 {t}^{2} - t + 2 = 0 \\ {t}^{2} (t - 2) - (t - 2) = 0 \\ (t - 2)( {t}^{2} - 1) = 0 \\ t1 = 2 \\ t2 = 1 \\ t3 = - 1$

корень -1 не подходит.

${2}^{x} = 1 \\ x1 = 0 \\ {2}^{x} = 2 \\ x2 = 1$

ответ: 0;1.