Question

(2/3)^3x-2 . (4/9)^2x+1 ≤ (8/27)^x-3

Answer 1

${( \frac{2}{3} )}^{3x - 2} . {( \frac{4}{9} )}^{2x + 1} \leqslant {( \frac{8}{27} )}^{x - 3}$

${( \frac{2}{3} )}^{3x - 2}. { (({\frac{2}{3} )}^{2} )}^{2x + 1} \leqslant {( {( \frac{2}{3} )}^{3} )}^{x - 3}$

${( \frac{2}{3} )}^{3x - 2}. {( \frac{2}{3} )}^{4x+2} \leqslant {( \frac{2}{3} )}^{3x - 9}$

${( \frac{2}{3} )}^{3x - 2 + 4x + 2} \leqslant {( \frac{2}{3} )}^{3x - 9}$

${ (\frac{2}{3}) }^{7x} \leqslant {( \frac{2}{3} )}^{3x - 9}$

Como a base está entre 0 e 1,o sentido da desigualdade se inverte. Daí

$7x \geqslant 3x - 9 \\ 7x - 3x \geqslant - 9 \\ 4x \geqslant - 9 \\ x \geqslant - \frac{9}{4}$