Question

2) Resolva em R, a inequação exponenecial​

Answer 1

Resposta:  

$x \le -\dfrac{9}{4}$

Explicação passo-a-passo:

Aplicaremos logaritmo na base 2/3, pois as bases são todas potências deste, então conseguiremos eliminá-lo das contas, de forma que:

$\left(\dfrac{2}{3} \right)^{3x-2}\cdot \left(\dfrac{4}{9} \right)^{2x+1} \le \left(\dfrac{8}{27} \right)^{x-3} \Leftrightarrow$

$Log_{\dfrac{2}{3}} \left[\left(\dfrac{2}{3} \right)^{3x-2}\cdot \left(\dfrac{4}{9} \right)^{2x+1} \right] \le Log_{\dfrac{2}{3}} \left(\dfrac{8}{27} \right)^{x-3} \Leftrightarrow$

$Log_{\dfrac{2}{3}} \left(\dfrac{2}{3} \right)^{3x-2} + Log_{\dfrac{2}{3}} \left(\dfrac{4}{9} \right)^{2x+1} \le Log_{\dfrac{2}{3}} \left(\dfrac{8}{27} \right)^{x-3} \Leftrightarrow$

$(3x-2)Log_{\dfrac{2}{3}} \left(\dfrac{2}{3} \right) + (2x+1)Log_{\dfrac{2}{3}} \left(\dfrac{4}{9} \right) \le (x-3)Log_{\dfrac{2}{3}} \left(\dfrac{8}{27} \right) \Leftrightarrow$

$(3x-2)\cdot1 + (2x+1)\cdot2 \le (x-3)\cdot3 \Leftrightarrow$

$(3x-2) + (4x+2) \le (3x-9) \Leftrightarrow 7x \le 3x-9 \Leftrightarrow 9 \le -4x \Leftrightarrow$

$x \le -\dfrac{9}{4} ~~~ \blacksquare$

Espero ter ajudado, tenha um bom dia!