e) 2²x -6.2* +8=0.

е) $$ 2^{2x} - 6 \cdot 2^x + 8 = 0 $$ Пусть $$ t = 2^x $$, тогда уравнение примет вид: $$ t^2 - 6t + 8 = 0 $$ $$ D = (-6)^2 - 4 \cdot 1 \cdot 8 = 36 - 32 = 4 $$ $$ t_1 = \frac{6 + \sqrt{4}}{2} = \frac{6 + 2}{2} = 4 $$ $$ t_2 = \frac{6 - \sqrt{4}}{2} = \frac{6 - 2}{2} = 2 $$ 1) $$ 2^x = 4 $$ $$ 2^x = 2^2 $$ $$ x_1 = 2 $$ 2) $$ 2^x = 2 $$ $$ x_2 = 1 $$ Ответ: x = 1; x = 2