Question

помогите пж очень нужно
Log2(2x-1)+log2(x+3)=2​

Answer 1

Объяснение:

$log_{2}(2x - 1) + log_{2}(x + 3 ) = 2 \\ xe < \frac{1}{2} . + \infty > \\ log_{2}((2x - 1) \times (x + 3)) = 2 \\ log_{8}( 2{x}^{2} + 6x - x - 3 ) = 2 \\ 2 {x}^{2} + 6x - x - 3 = {2}^{2} \\ 2 {x}^{2} + 5x - 3 = 4 \\ 2 {x}^{2} + 5x - 3 - 4 = 0 \\ 2 {x}^{2} + 7x - 2x - 7 = 0 \\ x(2x + 7) - (2x + 7) = 0 \\ (2x + 7)(x - 1) = 0 \\ 2x + 7 = 0 \\ x - 1 = 0 \\ x = - \frac{7}{2} \\ xe < \frac{1}{2} . + \infty > \\ x = 1$