Question

3- resolva a equação log de (x^2 + 2x - 7) na base 2 - log de (x - 1) na base 2 = 2

Answer 1

Utilizando a propriedade:   $log_{_b}\frac{a}{c}=log_{_b}a-log_{_b}c$

$log_{_2}(x^2+2x-7)-log_{_2}(x-1)=2\\\\log_{_2}\frac{x^2+2x-7}{x-1}=2\\\\\frac{x^2+2x-7}{x-1}=2^2\\\\x^2+2x-7=4\;.\;(x-1)\\\\x^2+2x-7=4x-4\\\\x^2-2x-3=0\\\\Bhaskara\\\\\Delta=(-2)^2-4.1.(-3)\\\\\Delta=4+12\\\\\Delta=16\\\\x_1=\frac{2+\sqrt{16}}{2.1}=\frac{2+4}{2}=3\\\\x_2=\frac{2-\sqrt{16}}{2.1}=\frac{2-4}{2}=-1$

Não podemos ter logaritmando negativo ou zero, logo x2 deve ser descartado. O valor de "x" fica então valendo x1.

Resp: x = 3