Решите уравнения 1. 2^(3x+2) = 8

1. $$2^{3x+2} = 8$$ $$2^{3x+2} = 2^3$$ $$3x+2 = 3$$ $$3x = 1$$ $$x = \frac{1}{3}$$ **Ответ:** $$\frac{1}{3}$$ 2. $$3^{x-6} = \frac{1}{9}$$ $$3^{x-6} = 3^{-2}$$ $$x-6 = -2$$ $$x = 4$$ **Ответ:** $$4$$ 3. $$5^{-x-2} = 125$$ $$5^{-x-2} = 5^3$$ $$-x-2 = 3$$ $$-x = 5$$ $$x = -5$$ **Ответ:** $$-5$$ 4. $$\left(\frac{1}{2}\right)^{4x-7} = 16$$ $$(2^{-1})^{4x-7} = 2^4$$ $$2^{-(4x-7)} = 2^4$$ $$-4x+7 = 4$$ $$-4x = -3$$ $$x = \frac{3}{4}$$ **Ответ:** $$\frac{3}{4}$$ 5. $$81^{5-x} = \frac{1}{3}$$ $$(3^4)^{5-x} = 3^{-1}$$ $$3^{4(5-x)} = 3^{-1}$$ $$4(5-x) = -1$$ $$20-4x = -1$$ $$-4x = -21$$ $$x = \frac{21}{4}$$ **Ответ:** $$\frac{21}{4}$$ 6. $$5 \cdot 25^x = 125$$ $$5^1 \cdot (5^2)^x = 5^3$$ $$5^1 \cdot 5^{2x} = 5^3$$ $$5^{1+2x} = 5^3$$ $$1+2x = 3$$ $$2x = 2$$ $$x = 1$$ **Ответ:** $$1$$ 7. $$(0,5)^{x^2-3} = 4$$ $$\left(\frac{1}{2}\right)^{x^2-3} = 2^2$$ $$(2^{-1})^{x^2-3} = 2^2$$ $$2^{-(x^2-3)} = 2^2$$ $$-x^2+3 = 2$$ $$-x^2 = -1$$ $$x^2 = 1$$ $$x = \pm 1$$ **Ответ:** $$\pm 1$$ 8. $$\left(\frac{1}{4}\right)^{x-5} = 256^x$$ $$(4^{-1})^{x-5} = (4^4)^x$$ $$4^{-(x-5)} = 4^{4x}$$ $$-x+5 = 4x$$ $$5 = 5x$$ $$x = 1$$ **Ответ:** $$1$$ 9. $$2^{3+x} = 0,4 \cdot 5^{3+x}$$ $$\frac{2^{3+x}}{5^{3+x}} = 0,4$$ $$\left(\frac{2}{5}\right)^{3+x} = \frac{4}{10}$$ $$\left(\frac{2}{5}\right)^{3+x} = \frac{2}{5}$$ $$3+x = 1$$ $$x = -2$$ **Ответ:** $$-2$$ 10. $$3^{x-2} - 3^{x-3} = 18$$ $$3^{x-3}(3^1 - 1) = 18$$ $$3^{x-3}(2) = 18$$ $$3^{x-3} = 9$$ $$3^{x-3} = 3^2$$ $$x-3 = 2$$ $$x = 5$$ **Ответ:** $$5$$ 11. $$7^{x-5} = 3^{x-5}$$ $$\frac{7^{x-5}}{3^{x-5}} = 1$$ $$\left(\frac{7}{3}\right)^{x-5} = \left(\frac{7}{3}\right)^0$$ $$x-5 = 0$$ $$x = 5$$ **Ответ:** $$5$$ 12. $$2^{2x} - 12 \cdot 2^x + 32 = 0$$ Пусть $$t = 2^x$$, тогда $$t^2 - 12t + 32 = 0$$ По теореме Виета или через дискриминант найдем корни: $$t_1 = 4, t_2 = 8$$ Для $$t_1 = 4$$ $$2^x = 4$$ $$2^x = 2^2$$ $$x = 2$$ Для $$t_2 = 8$$ $$2^x = 8$$ $$2^x = 2^3$$ $$x = 3$$ **Ответ:** $$2, 3$$ 13. $$9^x - 4 \cdot 3^x + 3 = 0$$ $$(3^2)^x - 4 \cdot 3^x + 3 = 0$$ $$(3^x)^2 - 4 \cdot 3^x + 3 = 0$$ Пусть $$t = 3^x$$, тогда $$t^2 - 4t + 3 = 0$$ По теореме Виета или через дискриминант найдем корни: $$t_1 = 1, t_2 = 3$$ Для $$t_1 = 1$$ $$3^x = 1$$ $$3^x = 3^0$$ $$x = 0$$ Для $$t_2 = 3$$ $$3^x = 3$$ $$3^x = 3^1$$ $$x = 1$$ **Ответ:** $$0, 1$$