Решите уравнение: 1) 5ˣ = 625; 2) 4³ˣ⁻¹⁷ = 64; 3) (1/2)²⁻³ˣ = 4ˣ⁺⁷; 4) 5ˣ⁺¹ + 5ˣ⁻¹ = 130; 5) 9ˣ - 7·3ˣ - 18 = 0; 6) 20ˣ²+³ˣ⁻⁴ = 1.

Решение: 1) $$5^x = 625$$ $$5^x = 5^4$$ $$x = 4$$ 2) $$4^{3x-17} = 64$$ $$4^{3x-17} = 4^3$$ $$3x - 17 = 3$$ $$3x = 20$$ $$x = \frac{20}{3}$$ 3) $$(\frac{1}{2})^{2-3x} = 4^{x+7}$$ $$(2^{-1})^{2-3x} = (2^2)^{x+7}$$ $$2^{-2+3x} = 2^{2x+14}$$ $$-2 + 3x = 2x + 14$$ $$x = 16$$ 4) $$5^{x+1} + 5^{x-1} = 130$$ $$5^x \cdot 5^1 + 5^x \cdot 5^{-1} = 130$$ $$5^x(5 + \frac{1}{5}) = 130$$ $$5^x(\frac{26}{5}) = 130$$ $$5^x = 130 \cdot \frac{5}{26}$$ $$5^x = 5 \cdot 5$$ $$5^x = 5^2$$ $$x = 2$$ 5) $$9^x - 7 \cdot 3^x - 18 = 0$$ $$(3^2)^x - 7 \cdot 3^x - 18 = 0$$ $$(3^x)^2 - 7 \cdot 3^x - 18 = 0$$ Пусть $$t = 3^x$$, тогда $$t^2 - 7t - 18 = 0$$ $$D = (-7)^2 - 4 \cdot 1 \cdot (-18) = 49 + 72 = 121$$ $$t_1 = \frac{7 + \sqrt{121}}{2} = \frac{7 + 11}{2} = \frac{18}{2} = 9$$ $$t_2 = \frac{7 - \sqrt{121}}{2} = \frac{7 - 11}{2} = \frac{-4}{2} = -2$$ Так как $$t = 3^x$$, то $$3^x = 9$$ или $$3^x = -2$$. Второе уравнение не имеет решений, т.к. $$3^x > 0$$ при любом x. $$3^x = 9$$ $$3^x = 3^2$$ $$x = 2$$ 6) $$20^{x^2+3x-4} = 1$$ $$20^{x^2+3x-4} = 20^0$$ $$x^2 + 3x - 4 = 0$$ $$D = 3^2 - 4 \cdot 1 \cdot (-4) = 9 + 16 = 25$$ $$x_1 = \frac{-3 + \sqrt{25}}{2} = \frac{-3 + 5}{2} = \frac{2}{2} = 1$$ $$x_2 = \frac{-3 - \sqrt{25}}{2} = \frac{-3 - 5}{2} = \frac{-8}{2} = -4$$ Ответы: 1) $$x = $$ extbf{4} 2) $$x = $$ extbf{20/3} 3) $$x = $$ extbf{16} 4) $$x = $$ extbf{2} 5) $$x = $$ extbf{2} 6) $$x_1 = $$ extbf{1}, $$x_2 = $$ extbf{-4}