Помогите пожалуйста: 1396 (а б) 1397 аб 1398 бв

1396а
$2^x \geq 4 \\\ 2^x \geq 2^2 \\\ x \geq 2$

1396б
$2^x\ \textless \ \frac{1}{2} \\\ 2^x\ \textless \ 2^{-1} \\\ x\ \textless \ -1$

1397а
$3^x \leq 81 \\\ 3^x \leq 3^4 \\\ x \leq 4$

1397б
$( \frac{1}{3} )^x\ \textgreater \ \frac{1}{27} \\\ ( \frac{1}{3} )^x\ \textgreater \ ( \frac{1}{3} )^3 \\\ x\ \textless \ 3$

1398б
$( \frac{2}{3} )^{3x+6}\ \textgreater \ \frac{4}{9} \\\ ( \frac{2}{3} )^{3x+6}\ \textgreater \ ( \frac{2}{3} )^2 \\\ 3x+6\ \textless \ 2 \\\ 3x\ \textless \ -4 \\\ x\ \textless \ - \frac{4}{3}$

1398в
$5^{4x+2} \geq 125 \\\ 5^{4x+2} \geq 5^3 \\\ 4x+2 \geq 3 \\\ 4x \geq 1 \\\ x \geq \frac{1}{4}$