Question

oi
Ajude me port favor

resolver:

A) (2ln x-1 ) (lnx - 3 ) = 0

B) (lnx)² +4lnx = 0

​​​

Answer 1

Explicação passo-a-passo:

a)

$\sf (2\cdot ln~x-1)\cdot(ln~x-3)=0$

Seja $\sf ln~x=y$

$\sf (2\cdot y-1)\cdot(y-3)=0$

• $\sf 2y-1=0$

$\sf 2y=1$

$\sf y'=\dfrac{1}{2}$

• $\sf y-3=0$

$\sf y"=3$

=> Para $\sf y=\dfrac{1}{2}$:

$\sf ln~x=y$

$\sf ln~x=\dfrac{1}{2}$

$\sf log_{e}~x=\dfrac{1}{2}$

$\sf x=e^{\frac{1}{2}}$

$\sf \red{x'=\sqrt{e}}$

=> Para $\sf y=3$

$\sf ln~x=y$

$\sf ln~x=3$

$\sf log_{e}~x=3$

$\sf \red{x"=e^3}$

O conjunto solução é:

$\sf S=\{\sqrt{e},e^3\}$

b)

$\sf (ln~x)^2+4\cdot ln~x=0$

Seja $\sf ln~x=y$

$\sf y^2+4y=0$

$\sf y\cdot(y+4)=0$

• $\sf y'=0$

• $\sf y+4=0$

$\sf y"=-4$

=> Para $\sf y'=0$:

$\sf ln~x=y$

$\sf ln~x=0$

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

$\sf x=e^0$

$\sf \red{x"=1}$

=> Para $\sf y"=-4$:

$\sf ln~x=y$

$\sf ln~x=-4$

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

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

$\sf \red{x"=\dfrac{1}{e^4}}$

O conjunto solução é:

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