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.

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