[tex] log(x) + log(x + 4) = log(5) \\ using \: formula \\ log(a) + log(b) = log(ab) we \: \: have \\ log(x)(x + 4) = log(5) \\ = > log( {x}^{2} + 4x) = log(5) \\ = > {x}^{2} + 4x = 5 \\ = > {x}^{2} + 4x - 5 = 0 \\ = > {x}^{2} + 5x - x - 5 = 0 \\ = > x(x + 5) - 1(x + 5) = 0 \\ = > (x - 1)(x + 5) = 0 \\ = > x = 1 \: \: or \: \: - 5 \\ but \: \: \: we \: \: \: cannot \: \: \: have \: \: \: negative \: \: \: numbers \\ so \: \: \: \: x \: \: = 1[/tex]
x = 1
