Find $x$ : $3 (2^{x} + 1) - 2^{x + 2} + 5 = 0$

Open in App

Solution

Consider,

$3 (2^{x} + 1) - 2^{x + 2} + 5 = 0$

$3 . 2^{x} + 3 - 2^{2} . 2^{x} + 5 = 0$ ------Product law $a^{m} . a^{n} = a^{m + n}$

$3 . 2^{x} + 3 - 4 . 2^{x} + 5 = 0$

$(3 . 2^{x} - 4 . 2^{x}) + (5 + 3) = 0$

$- 2^{x} + 8 = 0$

$2^{x} = 8 = 2^{3}$

$x = 3$ ---base is same, then the powers can be equated

Suggest Corrections

Similar questions

x

3 (2^{x} + 1) - 2^{x + 2} + 5 = 0

3 (2^{x} + 1) - 2^{x + 2} + 5 = 0

\frac{2 x}{x - 3} + \frac{1}{2 x + 3} + \frac{3 x + 9}{(x - 3) (2 x + 3)} = 0

x + 1

\frac{2}{x + 1} + \frac{3}{2 (x - 2)} = \frac{23}{5 x}; x \neq 0, - 1, 2

lim x \to 0 \frac{64^{x} - 32^{x} - 16^{x} + 4^{x} + 2^{x} - 1}{[\sqrt{(15 + cos x) - 4}] sin x} = - a {(log b)}^{3} .

a + b

Join BYJU'S Learning Program