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

We have the conditions : x + 3 > 0  and x - 2 > 0 ;
log2 ( x + 3 ) = log2 [ 3 / ( x - 2 ) ] ;
x + 3 = 3 / ( x - 2 ) ;
( x + 3 )( x - 2 ) = 3 ;
x^2 + 3x - 2x - 6 = 3 ;
x^2 + x - 9 = 0 ;
 For ax^2 + bx + c = 0, the value of x is given by:    ;
a = 1 ; b = 1 ; c = - 9 ;
then, x 1 ≈ 2.54 ; x 2 ≈ - 3.54 ;
The solutions are 2.54 and - 3.54 because its verify the conditions.

put like terms together.orremember that the 45 becomes negative when moved to the other side.orthe minuses cancel out.

We need more info then that