Question

Решите уравнение, плизз

Answer 1

Ответ:

1.

${( \frac{1}{7}) }^{x} = 49 \\ {7}^{ - x} = {7}^{2} \\ - x = 2 \\ x = - 2$

2.

${7}^{x} = 49 \\ {7}^{x} = {7}^{2} \\ x = 2$

3.

${7}^{x} = 1 \\ x = 0$

4.

${7}^{5x - 2} = {7}^{4x + 3} \\ 5x - 2 = 4x + 3 \\ x = 5$

5.

${7}^{15x + 3} = {49}^{x - 1} \\ {7}^{15x + 3} = {7}^{2( x - 1)} \\ 15x + 3 = 2x - 2 \\ 13x = - 5 \\ x = - \frac{5}{13}$

6.

$\frac{3}{2} \times {( \frac{1}{2} )}^{x} + ( \frac{1}{2} ) {}^{x} = \frac{5}{16} \\ \frac{5}{2} \times ( \frac{1}{2} ) {}^{x} = \frac{5}{16} \\ ( \frac{1}{2} ) {}^{x} = \frac{5}{16} \times \frac{2}{5} \\ ( \frac{1}{2} ) {}^{x} = \frac{1}{4} \\ x = 2$

7.

$4 \times ( \frac{1}{4} ) {}^{x} - 10 \times (\frac{1}{2} ) {}^{x} + 4 = 0 \\ 4 \times ( \frac{1}{2} ) {}^{2x} - 10 \times ( \frac{1}{2} ) {}^{x} + 4 = 0 \\ \\ (\frac{1}{2} ) {}^{x} = t \\ \\ 4t {}^{2} - 10 t + 4 = 0 \\ 2 {t}^{2} - 5t + 2 = 0 \\ D= 25 - 16 = 9 \\ t_1 = \frac{5 + 3}{4} = 2 \\ t_2 = \frac{1}{2} \\ \\ ( \frac{1}{2} ) {}^{x} = 2 \\ x_1 = - 1 \\ \\ ( \frac{1}{2} ) {}^{x} = \frac{1}{2} \\ x_2 = 1$