Question

2) log3 (x-1) + log3 (2x+1) – log3 (x-3) = 3

Answer 1

Olá, Raquel.

$\log_3 (x-1) + \log_3 (2x+1) -\log_3 (x-3) = 3\Rightarrow\\\\ \log_3 \frac{(x-1)(2x+1)}{(x-3)} = 3\Rightarrow\\\\ \frac{(x-1)(2x+1)}{(x-3)} = 3^3\Rightarrow\\\\ 2x^2+x-2x-1=27(x-3)\Rightarrow\\\\ 2x^2-x-1=27x-81\Rightarrow\\\\ 2x^2-28x+80=0\Rightarrow\\\\ x^2-14x+40=0$

As raízes desta equação são tais que:

$\begin{cases} x_1+x_2=14\\ x_1\cdot x_2=40 \end{cases}$

Ou seja:

$\boxed{x_1=10}\text{ e }\boxed{x_2=4}$