Question

Алгебра, решить уравнения, даю 10 баллов
$log_{3} {x}^{2} - log_{3} \frac{x}{x + 6} = 3$
$log_{3}x - 6 log_{2}3 = 1$
$log_{ {x}^{2} }16 - log_{ \sqrt{x} }7 = 2$
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Answer 1

log(3) x^2 - log(3) x/(x + 6) = 3

одз

x > 0

x/(x + 6) > 0    x < - 6   x > 0

x ∈ (0, +∞)

log(3) x^2 : x/(x + 6) = log(3) 3^3

log(3) x(x + 6) = log(3) 27

x^2 + 6x - 27 = 0

D = 36 + 4*27 = 144

x12 = (-6 +- 12)/2 = -9   3

x = -9 < 0 не корень

х = 3

=======

log(3) x - 6log(2) 3 = 1

одз x > 0

log(3) x = 1 + log(2) 3^6

log(3) x = log(2) 2*3^6

x = 3^(log(2) 2*3^6)

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log(x^2) 16 - log(√x) 7 = 2

одз x > 0  x ≠ 1 x ≠ - 1

x ∈ (0, 1) U (1, +∞)

1/2 log(x) 16 - 2 log(x) 7 = 2

log(x) 4 - log(x) 49 = log(x) x^2

log(x) 4/49 = log(x) x^2

x^2 = 4/49

x1 = -2/7 нет < 0

x2 = 2/7