Question

2) Determine a sabendo que: log 3a (8a+1) = 2​

Answer 1

Resposta:

Explicação passo a passo:

Determine a sabendo que:

log 3a (8a+1) = 2​

8a+1 = 3a^2

0 = 3a^2 - 8a - 1

3a^2 - 8a - 1 = 0

a = 3; b = - 8; c = - 1

a = x

∆= b^2 - 4ac

∆= (-8)^2 - 4.3.(-1)

∆= 64+12

∆= 76

√∆= √4.√19

√∆= 2√19

X = (-b +/- √∆]/2a

X = [-(-8)+/- 2√19]/2.3

X = (8 +/- 2√19)/6

X' = 8/6 + 2√19/6

X' = 4/3 + √19/3

X " = (8 - 2√19)/6

X" = 4/3 - √19/3

X = a

a ' = 4/3 + √19/3

a ' = 4/3 + 4,36/3

a ' = (8,36)/3

a' = 2,79

a = 4/3 - √19/3 (negativo): não serve

a " = 4/3 - 4,36/3 = (-0,36)/3

= - 0,12

Resp.:

a = 4/3 + √19/3

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{log_{3a}\:(8a + 1) = 2}$

$\mathsf{(3a)^2 = 8a + 1}$

$\mathsf{9a^2 = 8a + 1}$

$\mathsf{9a^2 - 8a - 1 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-8)^2 - 4.9.(-1)}$

$\mathsf{\Delta = 64 + 36}$

$\mathsf{\Delta = 100}$

$\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{8 \pm \sqrt{100}}{18} \rightarrow \begin{cases}\mathsf{x' = \dfrac{8 + 10}{18} = \dfrac{18}{18} = 1}\\\\\mathsf{x'' = \dfrac{8 - 18}{18} = -\dfrac{10}{18} = -\dfrac{5}{9}}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \{1\}}}}$