Question

Решите неравенство
$2{}^{4x + 3} > \frac{1}{16}$
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Answer 1

${2}^{4x + 3} = {2}^{3 } \\ 4x + 3 = 3 \\ x = 0$

Answer 2

$\displaystyle\\2^{4x+3}>\frac{1}{16} ~~\Leftrightarrow~~2^{4x+3}>2^{-4}\\\\2>1\\\\4x+3>-4;~4x>-7;~~4x>-\frac{7}{4} \\\\\\Otvet:x\in(-1,75;+\infty)$