In the equation $4^{x + 2} = 2^{x + 3} + 48$ , find the value of $x$ .

Open in App

Solution

Given $4^{x + 2} = 2^{x + 3} + 48$
$⟹ 2^{2 (x + 2)} = 2^{x + 3} + 48$
$⟹ 16 (2^{2 x}) - 8 (2^{x}) - 48 = 0$
$⟹ 16 (2^{x})^{2} - 32 (2^{x}) + 24 (2^{x}) - 48 = 0$
$⟹ (2^{x} - 2) (16 (2^{x}) + 24) = 0$
$⟹ 2^{x} = 2, - \frac{24}{16}$
$⟹ x = 1$

Suggest Corrections

Similar questions

$Find the value of a^{3} + a^{2} + a + 1, where 'a' satisfies the equation : \frac{4 x + 3}{2 x + 7} = \frac{2 x - 7}{x + 2}$

x

2^{x + 3} = 4^{x - 1}

2 x^{2} - x + 9 = x^{2} + 4 x + 3, x = 2, x = 3.

x

4 x - 3 = 2 x + 5

^{'} a^{'}

4 x^{2} + a x + 9 = {(2 x + 3)}^{2}

Join BYJU'S Learning Program