Question

Could you solve it? I don't know how to solve it:(

Answer 1

Answer:

-1, 4

Step-by-step explanation:

$log_2x+ log_2(x-3)=2 \\ log_2 \{x(x-3) \}=2 \\ x(x - 3) = {2}^{2} \\ {x}^{2} - 3x = 4 \\ {x}^{2} - 3x - 4 = 0 \\ {x}^{2} - 4x + x - 4 = 0 \\ x(x - 4) + 1(x - 4) = 0 \\ (x - 4)(x + 1) = 0 \\ x - 4 = 0 \: \: or \: \: x + 1 = 0 \\ x = 4 \: \: or \: \: x = - 1 \\ \huge \red{ \boxed{x = \{ - 1, \: \: 4 \}}}$