Logarithm question - spurious solution So I have the equation that I want to solve

$$\log_2(8x) - \log_2(1+\sqrt{x}) = 3, \hspace{1mm} x>0$$

My solution is $$\log_2\left(\frac{8x}{1+\sqrt{x}}\right) = 3 \\ \Rightarrow x = 1 + \sqrt{x} \\ \Rightarrow x - \sqrt{x} - 1 = 0 \\ \Rightarrow \sqrt{x} = \frac{1}{2} \pm \frac{\sqrt{5}}{2} \\ \Rightarrow x = \frac{3}{2} \pm \frac{\sqrt{5}}{2} $$ Now only the larger solution actually solves the equation. 
Why is this? I think it involves the squaring stage, but still a bit not sure

$$ \sqrt{x} = \frac{1}{2} \pm \frac{\sqrt{5}}{2} $$
At this stage you must ditch the negative solution  $\frac{1}{2} - \frac{\sqrt{5}}{2}$ beacause $\sqrt{x}$ is non negative.
