Solve the equation ${\log _2}(x + 3) + {\log _2}(x - 3) = 4$.

${\log _2}(x + 3) + {\log _2}(x - 3) = 4$

${\log _2}({x^2} - 9) = 4$     (M1)

${x^2} - 9 = {2^4}{\text{ }}( = 16)$     M1A1

${x^2} = 25$

$x =  \pm 5$     (A1)

$x = 5$     A1

[5 marks]