Phương trình log2(x-3)+log2(x-1)=3 có nghiệm là:

$$\log_2(x-3) + \log_2(x-1) = 3$$

TXĐ: $\textbf{D}=(3, \infty)$

$⇔\log_2[(x-3)(x-1)] = 3$

$⇔(x-3)(x-1) = 2^3$

$⇔x^2 - 4x + 3 = 8$

$⇔x^2 - 4x - 5 = 0$

$⇔ \left[ \begin{array}{l}x=5(t/m)\\x=-1(L)\end{array} \right.$

$@thomasnguyen1254$