Question

1) Resolva as seguintes equações .

2) Efetue as seguintes operações.
a) √ + √ b) √ . √ c) √ : √2

Answer 1

Explicação passo-a-passo:

a)

$\sf log_{4}~3+log_{4}~(x+2)=2$

$\sf log_{4}~[3\cdot(x+2)]=2$

$\sf 3\cdot(x+2)=4^2$

$\sf 3x+6=16$

$\sf 3x=16-6$

$\sf 3x=10$

$\sf x=\red{\dfrac{10}{3}}$

b)

$\sf log~2+log~(x+3)=log~16$

$\sf log~[2\cdot(x+3)]=log~16$

Igualando os logaritmandos:

$\sf 2\cdot(x+3)=16$

$\sf 2x+6=16$

$\sf 2x=16-6$

$\sf 2x=10$

$\sf x=\dfrac{10}{2}$

$\sf \red{x=5}$

c)

$\sf log~(2x+3)-log~3=log~7$

$\sf log~\left(\dfrac{2x+3}{3}\right)=log~7$

Igualando os logaritmandos:

$\sf \dfrac{2x+3}{3}=7$

$\sf 2x+3=3\cdot7$

$\sf 2x+3=21$

$\sf 2x=21-3$

$\sf 2x=18$

$\sf x=\dfrac{18}{2}$

$\sf \red{x=9}$

d)

$\sf log~(x+8)-log~2=log~(2x+3)$

$\sf log~\left(\dfrac{x+8}{2}\right)=log~(2x+3)$

Igualando os logaritmandos:

$\sf \dfrac{x+8}{2}=2x+3$

$\sf x+8=2\cdot(2x+3)$

$\sf x+8=4x+6$

$\sf 4x-x=8-6$

$\sf 3x=2$

$\sf \red{x=\dfrac{2}{3}}$