Solve the following equation $2^{2 x} + 2^{x + 2} - 32 = 0$ .

Open in App

Solution

$2^{x + 2} = 2^{x} . 2^{2} = {4.2}^{x} 2^{x} = t$
$\Rightarrow e q^{n} : t^{2} + 4 t - 32 = 0 \Rightarrow t^{2} + 8 t - 4 t - 32 = 0$
$\Rightarrow t (t + 8) - 4 (t + 8) = 0 \Rightarrow (t - 4) (t + 8) = 0$
$\Rightarrow t = 4 o r t = - 8 ⨂ (2^{x} > 0) \Rightarrow 2^{x} = 4 \Rightarrow x = 2$

Suggest Corrections

Similar questions

2^{2 x} - {3.2}^{(x + 2)} + 32 = 0

Solve each of the following quadratic equations:
$2^{2 x} - {3.2}^{(x + 2)} + 32 = 0$

2^{2 x} + 2^{x + 2} - 4 \times 2^{3} = 0

2^{2 x} - 2^{x + 3} + 2^{4} = 0

2 {(x - 3)}^{2} + 3 (x - 2) (2 x - 3) = 8 (x + 4) (x - 4) - 1

Join BYJU'S Learning Program