2) 3ˣ + 4⋅3⁻ˣ⁺¹ = 13;

2) $$ 3^x + 4 \cdot 3^{-x+1} = 13 $$

$$ 3^x + \frac{12}{3^x} = 13 $$

$$ (3^x)^2 + 12 = 13 \cdot 3^x $$

$$ (3^x)^2 - 13 \cdot 3^x + 12 = 0 $$

Пусть $$ t = 3^x $$, тогда

$$ t^2 - 13t + 12 = 0 $$

$$ D = 169 - 48 = 121 $$

$$ t_1 = \frac{13 + 11}{2} = 12, t_2 = \frac{13 - 11}{2} = 1 $$

1) $$ 3^x = 12, x = \log_3 12 $$

2) $$ 3^x = 1, x = 0 $$

Ответ: log₃12, 0