Question

log2 (x-3)+log2(2x+1)=2

Answer 1

Log2 (x-3)+log2(2x+1)=2

Log2 (x-3)+log2(2x+1)=2, x∈ [3, +∞]

log2 ((x-3) * (2x+1)) =2

log2 (2x^2 +x -6x -3)=2

2x^2 +x -6x -3=2^2

2x^2- 5x -3 =4

2x^2 -5x -3 -4 =0  

2x^2 -5x-7=0

x= -(-5) ± √-(-5)^2 -4 *2*(-7) / 2*2

x= 5± √25+56/ 4

x= 5 ±√81/4

x=5±9 / 4

x= 5+9/4                     x=5-9/4

x=7/2, x ∈ [3, + ∞]

x=-1

x=7/2