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.