3 Решить уравнение: 1) 3ˣ⁺¹ = 27ˣ⁻¹; 2) 0,2ˣ²+⁴ˣ⁻⁵ = 1; 3) 2ˣ⁺³ - 2ˣ⁺¹ = 12; 4) 4⋅2²ˣ - 5⋅2ˣ + 1 = 0.

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

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

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

$$ 2x = 4 $$

$$ x = 2 $$

2) $$ 0.2^{x^2+4x-5} = 1 $$

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

$$ (x+5)(x-1) = 0 $$

$$ x_1 = -5, x_2 = 1 $$

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

$$ 8 \cdot 2^x - 2 \cdot 2^x = 12 $$

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

$$ 2^x = 2 $$

$$ x = 1 $$

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

$$ 4t^2 - 5t + 1 = 0 $$

$$ D = 25 - 16 = 9 $$

$$ t_1 = \frac{5 + 3}{8} = 1, t_2 = \frac{5 - 3}{8} = \frac{1}{4} $$

1) $$ 2^x = 1, x = 0 $$

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

Ответ: 1) 2; 2) -5, 1; 3) 1; 4) 0, -2