Question

Ajude me por favor

resolver:

3 ln²x -8 lnx -3 = 0

ln³x -5 ln²x +4 lnx = 0​

Answer 1

Explicação passo-a-passo:

a)

$\sf 3\cdot~ln~x^2-8\cdot ln~x-6=0$

Seja $\sf ln~x=y$

$\sf 3y^2-8y-6=0$

$\sf \Delta=(-8)^2-4\cdot3\cdot(-6)$

$\sf \Delta=64+72$

$\sf \Delta=136$

$\sf y=\dfrac{-(-8)\pm\sqrt{136}}{2\cdot3}=\dfrac{8\pm2\sqrt{34}}{2}$

$\sf y'=\dfrac{8+2\sqrt{34}}{6}~\Rightarrow~y'=\dfrac{4+\sqrt{34}}{3}$

$\sf y"=\dfrac{8-2\sqrt{34}}{6}~\Rightarrow~y"=\dfrac{4-\sqrt{34}}{3}$

=> Para $\sf y'=\dfrac{4+\sqrt{34}}{3}$:

$\sf ln~x=y$

$\sf ln~x=\dfrac{4+\sqrt{34}}{3}$

$\sf log_{e}~x=\dfrac{4+\sqrt{34}}{3}$

$\sf x'=e^{\frac{4+\sqrt{34}}{3}}$

$\sf \red{x'=\sqrt[3]{e^{4+\sqrt{34}}}}$

=> Para $\sf y"=\dfrac{4-\sqrt{34}}{3}$:

$\sf ln~x=y$

$\sf ln~x=\dfrac{4-\sqrt{34}}{3}$

$\sf log_{e}~x=\dfrac{4-\sqrt{34}}{3}$

$\sf x'=e^{\frac{4-\sqrt{34}}{3}}$

$\sf \red{x'=\sqrt[3]{e^{4-\sqrt{34}}}}$

O conjunto solução é:

$\sf S=\{\sqrt[3]{e^{4-\sqrt{34}}},\sqrt[3]{e^{4+\sqrt{34}}}\}$

b)

$\sf ln^3~x-5\cdot ln^2~x+4\cdot ln~x=0$

Seja $\sf ln~x=y$

$\sf y^3-5y^2+4y=0$

$\sf y\cdot(y^2-5y+4)=0$

• $\sf y'=0$

• $\sf y^2-5y+4=0$

$\sf \Delta=(-5)^2-4\cdot1\cdot4$

$\sf \Delta=25-16$

$\sf \Delta=9$

$\sf y=\dfrac{-(-5)\pm\sqrt{9}}{2\cdot1}=\dfrac{5\pm3}{2}$

$\sf y'=\dfrac{5+3}{2}~\Rightarrow~y=\dfrac{8}{2}~\Rightarrow~\red{y"=4}$

$\sf y"=\dfrac{5-3}{2}~\Rightarrow~y"=\dfrac{2}{2}~\Rightarrow~\red{y'"=1}$

=> Para y = 0:

$\sf ln~x=y$

$\sf ln~x=0$

$\sf log_{e}~x=0$

$\sf x=e^0$

$\sf \red{x=1}$

=> Para y = 4:

$\sf ln~x=y$

$\sf ln~x=4$

$\sf log_{e}~x=4$

$\sf \red{x=e^4}$

=> Para y = 1:

$\sf ln~x=y$

$\sf ln~x=1$

$\sf log_{e}~x=1$

$\sf x=e^1$

$\sf \red{x=e}$

O conjunto solução é:

$\sf S=\{1,e,e^4\}$