4) 4ˣ + 2²ˣ⁺¹ - 80 = 0.

4) $$ 4^x + 2^{2x+1} - 80 = 0 $$

$$ (2^2)^x + 2 \cdot 2^{2x} - 80 = 0 $$

$$ (2^x)^2 + 2 \cdot (2^x)^2 - 80 = 0 $$

$$ 3 \cdot (2^x)^2 = 80 $$

$$ (2^x)^2 = \frac{80}{3} $$

$$ 2^x = \sqrt{\frac{80}{3}} $$

$$ x = \log_2 \sqrt{\frac{80}{3}} = \frac{1}{2} \log_2 \frac{80}{3} $$

Ответ: 0.5log₂(80/3)