Question

Resolva: 
a) log 6 x + log 6 (x+1)=1
b) log3 (x-2) + log 3(5x+2)=2+log3(2x-1)

Answer 1

a) Utilizando a propriedade dos logs.

$log\ _6\ x+log\ _6\ (x+1)=1\\\\ log\ _6\ x(x+1) = 1\\\\ 6^1= x(x+1)\\\\ 6 = x^2+x\\\\ x^2+x-6=0\\\\ \Delta = 25\\\\ x^1 = 2\\\\ x^2 = -3$

-3 não pode ser....


b) $log\ _3\ (x-2)+ log\ _3\ (5x+2) = 2+log\ _3(2x-1)\\\\ log\ _3\ (x-2)(5x+2) = 2+log\ _3(2x-1)\\\\ log\ _3\frac{(x-2)(5x+2)}{(2x-1)} = 2\\\\ 3^2 = \frac{(x-2)(5x+2)}{(2x-1)}\\\\ 18x-9 = 5x^2-8x-4\\\\ 5x^2-26x+5=0\\\\ \Delta = 576\\\\ x^1 = 5\\\\ x^2 = \frac{1}{5}$

1/5 não pode ser solução.