3 Решить уравнение: 1) 3^{x + 1} = 27^{x-1}; 2) 0,2^{x^2 + 4x - 5} = 1; 3) 2^{x + 3} - 2^{x+1} = 12; 4) 4 \cdot 2^{2x} - 5 \cdot 2^x + 1 = 0.

Решить уравнение:

1) $$3^{x + 1} = 27^{x-1};$$

$$3^{x + 1} = (3^3)^{x-1};$$

$$3^{x + 1} = 3^{3(x-1)};$$

$$x + 1 = 3(x - 1);$$

$$x + 1 = 3x - 3;$$

$$2x = 4;$$

$$x = 2$$.

2) $$0,2^{x^2 + 4x - 5} = 1;$$

$$0,2^{x^2 + 4x - 5} = 0,2^0;$$

$$x^2 + 4x - 5 = 0;$$

По теореме Виета:

$$x_1 + x_2 = -4;$$

$$x_1 \cdot x_2 = -5;$$

$$x_1 = 1;$$

$$x_2 = -5$$.

3) $$2^{x + 3} - 2^{x+1} = 12;$$

$$2^x \cdot 2^3 - 2^x \cdot 2^1 = 12;$$

$$2^x (8 - 2) = 12;$$

$$2^x \cdot 6 = 12;$$

$$2^x = 2;$$

$$x = 1$$.

4) $$4 \cdot 2^{2x} - 5 \cdot 2^x + 1 = 0.$$.

$$4 \cdot (2^{x})^2 - 5 \cdot 2^x + 1 = 0;$$

Пусть $$t = 2^x$$, тогда $$4t^2 - 5t + 1 = 0$$.

$$D = (-5)^2 - 4 \cdot 4 \cdot 1 = 25 - 16 = 9;$$

$$t_1 = \frac{5 + 3}{8} = 1;$$

$$t_2 = \frac{5 - 3}{8} = \frac{1}{4}$$.

1) $$2^x = 1;$$

$$x = 0$$.

2) $$2^x = \frac{1}{4};$$

$$2^x = 2^{-2};$$

$$x = -2$$.

Ответ: 1) x = 2; 2) x = 1, x = -5; 3) x = 1; 4) x = 0, x = -2.