Страница 80 номер 216, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (208—223). 1) \(10^x = \sqrt[3]{100}\); 2) \(10^x = \sqrt[5]{10\,000}\); 3) \(225^{2x^2 - 24} = 15\); 4) \(10^x = \frac{1}{\sqrt[4]{10\,000}}\); 5) \((\sqrt{10})^x = 10^{x^2 - x}\); 6) \(100^{x^2 - 1} = 10^{1 - 5x}\).
1) \(10^{x} = \sqrt[3]{100}\)
\(10^{x} = 10^{\frac{2}{3}},\)
\(x = \dfrac{2}{3};\)
2) \(10^{x} = \sqrt[5]{10000}\)
\(10^{x} = 10^{\frac{4}{5}}, x = \dfrac{4}{5};\)
3) \(225^{2x^{2} - 24} = 15\)
\(15^{2(2x^{2} - 24)} = 15^{1},\)
\(4x^{2} - 48 = 1,\)
\(4x^{2} = 49,\)
\(2x = \pm 7,\)
\(x = \pm 3{,}5;\)
4) \(10^{x} = \frac{1}{\sqrt[4]{10000}}\)
\(10^{x} = \dfrac{1}{10},\)
\(10^{x} = 10^{-1},\)
\(x = -1;\)
5) \(\left(\sqrt{10}\right)^{x} = 10^{x^{2} - x}\)
\(10^{\frac{x}{2}} = 10^{x^{2} - x},\)
\(\dfrac{x}{2} = x^{2} - x,\)
\(x = 2x^{2} - 2x,\)
\(x(2x - 3) = 0,\)
\(x_{1} = 0, x_{2} = \dfrac{3}{2};\)
6) \(100^{x^{2} - 1} = 10^{1 - 5x}\)
\(10^{2x^{2} - 2} = 10^{1 - 5x},\)
\(2x^{2} - 2 = 1 - 5x,\)
\(2x^{2} + 5x - 3 = 0,\)
\(x_{1{,}2} = \dfrac{-5 \pm \sqrt{25 + 24}}{4},\)
\(x_{1} = \dfrac{1}{2}, x_{2} = -3.\)
