Question

Inequações exponenciais 1° ANO OBS1: Acho que tem baskara nas duas ultimas, eu não sei ajudem me please OBS2: A primeira linha de cada questão ja ta respondida

Answer 1

Explicação passo-a-passo:

a)

$\sf 4^{x-7} < \left(\dfrac{1}{32}\right)^x$

$\sf (2^{2})^{x-7} < (2^{-5})^{x}$

$\sf 2^{2x-14} < 2^{-5x}$

$\sf 2x-14 < -5x$

$\sf 2x+5x < 14$

$\sf 7x < 14$

$\sf x < \dfrac{14}{7}$

$\sf x < 2$

$\sf \red{S=\{x\in\mathbb{R}~|~x < 2\}}$

b)

$\sf (\sqrt{3})^{2x+4}\ge\left(\dfrac{1}{9}\right)^{3x}$

$\sf (3^{\frac{1}{2}})^{2x+4}\ge(3^{-2})^{3x}$

$\sf 3^{\frac{2x+4}{2}}\ge3^{-6x}$

$\sf 3^{x+2}\ge3^{-6x}$

$\sf x+2\ge-6x$

$\sf x+6x\ge-2$

$\sf 7x\ge-2$

$\sf x\ge\dfrac{-2}{7}$

$\sf \red{S=\left\{x\in\mathbb{R}~|~x\ge\dfrac{-2}{7}\right\}}$

c)

$\sf \left(\dfrac{1}{3}\right)^{x+4}\le\left(\dfrac{1}{27}\right)^{x^2}$

$\sf (3^{-1})^{x+4}\le(3^{-3})^{x^2}$

$\sf 3^{-x-4}\le3^{-3x^2}$

$\sf -x-4\le-3x^2$

$\sf 3x^2-x-4\le0$

$\sf \Delta=(-1)^2-4\cdot3\cdot(-4)$

$\sf \Delta=1+48$

$\sf \Delta=49$

$\sf x=\dfrac{-(-1)\pm\sqrt{49}}{2\cdot3}=\dfrac{1\pm7}{6}$

$\sf x'=\dfrac{1+7}{6}~\rightarrow~x'=\dfrac{8}{6}~\rightarrow~x'=\dfrac{4}{3}$

$\sf x"=\dfrac{1-7}{6}~\rightarrow~x"=\dfrac{-6}{6}~\rightarrow~x"=-1$

$\sf \red{S=\left\{x\in\mathbb{R}~|~-1\le x\le\dfrac{4}{3}\right\}}$

d)

$\sf \left(\dfrac{1}{8}\right)^{x^2} > 64^{2x+2}$

$\sf (2^{-3})^{x^2} > (2^6)^{2x+2}$

$\sf 2^{-3x^2} > 2^{12x+12}$

$\sf -3x^2 > 12x+12$

$\sf 3x^2+12x+12 < 0$

$\sf x^2+4x+4 < 0$

$\sf \Delta=4^2-4\cdot1\cdot4$

$\sf \Delta=16-16$

$\sf \Delta=0$

$\sf x=\dfrac{-4\pm\sqrt{0}}{2\cdot1}=\dfrac{-4\pm0}{2}$

$\sf x'=x"=\dfrac{-4}{2}$

$\sf x'=x"=-2$

Não há solução real

$\sf \red{S=\{~\}}$

e)

$\sf \left(\dfrac{5}{4}\right)^{-x+2} > \dfrac{64}{125}$

$\sf \left[\left(\dfrac{4}{5}\right)^{-1}\right]^{-x+2} > \left(\dfrac{4}{5}\right)^3$

$\sf \left(\dfrac{4}{5}\right)^{x-2} > \left(\dfrac{4}{5}\right)^3$

$\sf x-2 > 3$

$\sf x > 3+2$

$\sf x > 5$

$\sf \red{S=\{x\in\mathbb{R}~|~x > 5\}}$