Question

A equação log₂ (x+1) + log₂⁡ (x+2) - log₂ (x-1)= log₂ ⁡10 possui duas raízes. A soma dessas raízes é igual a:

Answer 1

Vamos usar as seguintes propriedades de log :

$\displaystyle \sf \underline{Soma\ vira\ produto} : \\\\ \log_a x +\log_ay = log_a(x\cdot y ) \\\\\\ \underline{Diferen{\c c}a\ vira\ divis{\~a}o }} : \\\\ \log_ax-\log_ay = \log_a\left(\frac{x}{y} \right)$

Temos :

$\displaystyle \sf \underbrace{\sf \log_2(x+1)+\log_2(x+2)}_{\displaystyle \sf \log_2\left[(x+1)\cdot (x+2)\right]}-\log_2(x-1) = \log_210 \\\\\\ \log_2(x^2+3x+2)-\log_2(x-1)=\log_210\\\\\\ \log_2\left[\frac{x^2+3x+2}{x-1}\right]=\log_210 \\\\\\ \frac{x^2+3x+2}{x-1}=10 \to x^2+3x+2=10(x-1)\\\\\\ x^2+3x+2=10x-10 \\\\ x^2+3x-10x+2+10 = 0 \\\\ x^2-7x+12=0 \\\\ Soma\ das \ ra{\'i}zes} : \\\\ x_1+x_2 = \frac{-b}{a} \\\\ x_1+x_2 = \frac{-(-7)}{1} \\\\ \huge\boxed{\sf x_1+x_2 = 7 }\checkmark$