Question

O produto do conjunto solução da equação exponencial a seguir é

2 ^ (2x + 1) - 2 ^ (x + 4) - 2 ^ x + 8 = 0

a) 1

b) -3

c) 8

d) -1

e) 3​

Answer 1

$\displaystyle \sf 2^{2x+1}-2^{x+4}-2^x+8=0 \\\\ 2^{2x}\cdot 2^1 - 2 ^{x}\cdot 2^4 -2^x +8=0 \\\\ 2\cdot (2^x)^2 - 16\cdot (2^x )-(2^x)+8=0 \\\\ 2\cdot (2^x)^2 - 17\cdot (2^x )+8=0 \\\\ Fa{\c c}amos : \\\\ 2^x = y \\\\ Da{\'i}} : \\\\ 2y^2-17y+8=0\\\\ y = \frac{-(-17)\pm\sqrt{(-17)^2-4\cdot 2 \cdot 8 }}{2\cdot 2} \\\\\\ y = \frac{17\pm\sqrt{289-64}}{4} \to y = \frac{17\pm\sqrt{225}}{4}$

$\displaystyle \sf y = \frac{17\pm15}{4}\to \left\{\begin{array}{I}\displaystyle \sf y = \frac{17+15}{4}\to y = \frac{32}{4} \to \boxed{\sf y= 8 } \\\\\\ \displaystyle \sf y = \frac{17-15}{4}\to y = \frac{2}{4} \to \boxed{\sf y= \frac{1}{2} } \end{array} \right \\\\\\ Da{\'i}}, \ temos : \\\\\ 2^x = y \\\\\\ 2^x = 8 \to 2^x = 2^3 \to \boxed{\sf x = 3} \\\\\ 2^x = \frac{1}{2} \to 2^x = 2^{-1} \to \boxed{\sf x = -1 }$

$\displaystyle \sf S = \{ x = 3 \ e \ x = -1\} \\\\ \text{A quest{\~a}o pede o produto do conjunto solu{\c c}{\~a}o, ou seja} : \\\\ 3\cdot (-1) \\\\ \huge\boxed{-3 }\checkmark$

letra b