• Предмет: Математика
  • Автор: ississiiei
  • Вопрос задан 6 лет назад

решите пожалуйста уравнение​

Приложения:

Ответы

Ответ дал: TheEvilGenius
0

1.

log₄(log₃x) = 1

log₃x = 4¹

x = 3⁴ = 81

2.

lgx - lg5 = 2lgx - lg(9 - x)

lgx - 2lgx + lg(9 - x) = lg5

-lgx + lg(9 - x) = lg5

lg((9 - x) / x) = lg5

(9 - x) / x = 5

x = 1.5

3.

lgx = t

t² - t = 6

t₁ = -2  |  t₂ = 3

lgx₁ = -2

x₁ = 1 / e²

x₂ = e³

4.

frac{log_{10}16}{log_{10}x^2} + frac{log_{10}64}{log_{10}2x} = 3\frac{log_{10}2^4}{log_{10}x^2} + frac{log_{10}2^6}{log_{10}2x} = 3\frac{4log_{10}2}{log_{10}x^2} + frac{6log_{10}2}{log_{10}2x} = 3\frac{4log_{10}(2)log_{10}(2x) + 6log_{10}(2)log_{10}(x^2)}{log_{10}(x^2)log_{10}(2x)} = 3\4log_{10}(2)log_{10}(2x) + 6log_{10}(2)log_{10}(x^2) = 3log_{10}(x^2)log_{10}(2x)\4log_{10}(2)(log_{10}(2) + log_{10}(x)) + 12log_{10}(2)log_{10}(|x|) = 6log_{10}(|x|)log_{10}(2x)\

4log^2_{10}(2) + 4log_{10}(2)log_{10}(x)+ 12log_{10}(2)log_{10}(x) = 6log_{10}(x)log_{10}(2x)\-3log^2_{10}(x) + 5log_{10}(2)log_{10}(x) + 2log^2_{10}(2) = 0\t = log_{10}(x)\-3t^2 + 5log_{10}(2)t + 2log^2_{10}(2) = 0\t_1 = 2log_{10}(2)\t_2 = -frac{log_{10}(2)}{3}\log_{10}(x_1) = 2log_{10}(2)\x_1 = 4\log_{10}(x_2) = -frac{log_{10}(2)}{3}\x_2 = frac{sqrt[3]{4}}{2}

5.

log_{3^{-1}}(x^{log_{3^{-1}}(x)-frac{2}{3}}) = log_{3^{-1}}(3^{frac{8}{3}})\log_{3^{-1}}(x^{-log_{3}(x)-frac{2}{3}}) = -frac{8}{3}\(log_3(x) + frac{2}{3})log_3(x) = -frac{8}{3}\log^2_3(x) + frac{2}{3}log_3(x) = -frac{8}{3}\t = log_3(x)\t^2 + frac{2}{3}t = -frac{8}{3}\t notin R\x notin R

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