Question

Помогите решить, лагорифм и ур-ия

Answer 1

1.

$log_{2}( \frac{x + 3}{4x - 5} ) = 0 \\ \frac{x + 3}{4x - 5} = {2}^{0} \\ x + 3 = 4x - 5 \\ 3x = 8 \\ x = \frac{8}{3}$

2.

${ log_{3} x } = y \\ {y }^{2} - 5y + 6 = 0 \\ y1 + y2 = 5 \\ y1 \times y2 = 6 = > y1 = 2 \\ y2 = 3 \\ log_{3}(x) = 2 \\ x = {3}^{2} = 9 \\ log_{3}(x) = 3 \\ x = 27$

3.

${6}^{x} = y > 0 \\ {y}^{2} - 3y - 18 = 0 \\ y1 + y2 = 3 \\ y1 \times y2 = - 18 = > y1 = 6 \\ y2 = - 3 \\ {6}^{x} = 6 = > x = 1 \\ {6}^{x} = - 3(net)$

ответ : х = 1.

4.

$( { \frac{3}{7}) }^{3x - 7} = ( { \frac{3}{7}) }^{ - (7x - 3)} \\ 3x - 7 = - 7x + 3 \\ 10x = 10 \\ x = 1$