Question

БУДЬ-ЛАСКА я ДОПОМОЖІТЬ

Answer 1

Ответ:

1.

${16}^{x} = 8 \\ {2}^{4x} = {2}^{3} \\ 4x = 3 \\ x = \frac{3}{4}$

2.

${5}^{4x + {x}^{2} } = 1 \\ 4x + {x}^{2} = 0 \\ x(x + 4) = 0 \\ x1 = 0 \\ x2= - 4$

3.

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

4.

$( \frac{1}{2} ) {}^{x} < \frac{1}{16} \\ x > 4$

5.

$log_{4}(x) = 3 \\ x = {4}^{3} \\ x = 64$

6.

$log_{5}(3x - 1) = log_{5}( {x}^{2} - 1 ) \\ 3x - 1 = {x}^{2} - 1 \\ {x}^{2} - 3x = 0 \\ x(x - 3) = 0 \\ x1 = 0 \\ x2 = 3$

Корень 0 не подходит

Ответ: 3

7.

$log_{2}( {2}^{x} - 6 ) = x - 2 \\ {2}^{x} - 6 = {2}^{x - 2} \\ {2}^{x} - {2}^{x - 2} = 6 \\ {2}^{x - 2} ( {2}^{2} - 1) = 6 \\ {2}^{x - 2} = 2\\ x - 2 = 1 \\ x = 3$