Question

Помогите пж надо по шогам;((

Answer 1

Ответ:

$1) {12}^{x} - 6 \times {6}^{x} + 8 \times {3}^{x} = 0 \\ {4}^{x} \times {3}^{x} - 6 \times {2}^{x} \times {3}^{x} + 8 \times {3}^{x} = 0 \\ {3}^{x} ( {4}^{x} - 6 \times {2}^{x} + 8) = 0$

${3}^{x} = 0$

нет корней.

${2}^{2x} - 6 \times {2}^{x} + 8 = 0$

Замена:

${2}^{x} = t \\ {t}^{2} - 6t + 8 = 0 \\ (t - 2)(t - 4) = 0 \\ t1 = 2 \\ t2 = 4$

${2}^{x} = 2 \\ x1 = 1 \\ {2}^{x} = 4 \\ x2 = 2$

Ответ: 1,2.

$2) {5}^{x + 1} - {5}^{x - 1} = {4}^{x + 1} + 2 \times {4}^{x} \\ {5}^{x - 1} ( {5}^{2} - 1) = {4}^{x} (4 + 2) \\ {5}^{x - 1} \times 24 = {4}^{x} \times 6 \\ \frac{ {5}^{x - 1} }{ {4}^{x} } = \frac{6}{24} \\ \frac{ {5}^{x} \times {5}^{ - 1} }{ {4}^{x} } = \frac{1}{4} \\ {( \frac{5}{4}) }^{x} = \frac{5}{4} \\ x = 1$

Ответ: 1.