Question

Срочно!!!
$log_{0,7}(2x - 7) > log_{0,7} ^{5}$
$log_{3}(x - 2) + log_{3}(x + 6) \leqslant 2$
$log_{3} ^{2} {x} - log{3} ^{x < 0}$
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Answer 1

log(0.7) (2x - 7) > log(0.7) 5

одз  x > 7/2

Основание < 1 меняем знак неравенства

2x - 7 < 5

2x < 12

x < 6

x ∈ (7/2, 6)

------

log(3) (x - 2) + log(3) ( x + 6) ≤ 2

одз  x > 2  x > -6

x ∈ (2, +∞)

log(3) (x - 2)(x + 6) ≤ log(3) 3²

Основание > 1 не меняем знак неравенства

x² + 6x - 2x - 12 - 9 ≤ 0

x² + 4x - 21 ≤ 0

D = 16 + 84 = 100

x12 = (-4 +- 10)/2 = -7   3

(x - 3)(x + 7) ≤ 0

++++++[-7] -------------- [3] +++++++

x ∈ [-7,3]

+ одз

x ∈ (2,3]

-------

log²(3) x - log(3) x < 0

одз x > 0

log(3) x = y

y² - y < 0

y(y - 1) < 0

++++++(0) ------------ (1) +++++++++

y ∈ (0,1)

log(3) x > 0

x > 1

log(3) x < 1

x < 3

x ∈ (1,3)

+ одз

x ∈ (1,3)