Страница 80 номер 225, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (225—226). 1) \(3^{2x + 6} = 2^{x + 3}\); 2) \(5^{x - 2} = 4^{2x - 4}\); 3) \(2^x \cdot 3^x = 36^{x^2}\); 4) \(9^{-\sqrt{x - 1}} = \frac{1}{27}\).
1) \(3^{2x+6} = 2^{x+3}\)
\(3^{2(x+3)} = 2^{x+3},\)
\(\left(\dfrac{9}{2}\right)^{x+3} = \left(\dfrac{9}{2}\right)^{0},\)
\(x + 3 = 0,\)
\(x = -3;\)
2) \(5^{x-2} = 4^{2x-4}\)
\(5^{x-2} = 4^{2(x-2)},\)
\(\left(\dfrac{5}{16}\right)^{x-2} = \left(\dfrac{5}{16}\right)^{0},\)
\(x - 2 = 0,\)
\(x = 2;\)
3) \(2^{x} \cdot 3^{x} = 36^{x^{2}}\)
\(6^{x} = 6^{2x^{2}},\)
\(x = 2x^{2},\)
\(x(2x - 1) = 0,\)
\(x_{1} = 0;\)
\(2x = 1,\)
\(x_{2} = \dfrac{1}{2};\)
4) \(9^{-\sqrt{x-1}} = \frac{1}{27}\)
\(3^{-2\sqrt{x-1}} = 3^{-3},\)
\(-2\sqrt{x - 1} = -3,\)
\(\sqrt{x - 1} = \dfrac{3}{2},\)
\(x - 1 = \dfrac{9}{4},\)
\(x = \dfrac{13}{4} = 3\dfrac{1}{4}.\)
