Logarithm question - spurious solution

Question

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

Answer 1

The problem is in the penultimate step where you have $$\sqrt x=\frac{1-\sqrt 5}{2}.$$ This is simply false according to the accepted conventions of how to interpret $\sqrt{}.$

Moreover, note that there is a second solution since the equation $x-\sqrt x-1=0$ may also be written as $$(-\sqrt x)^2+(-\sqrt x)-1=0,$$ whence we find that $$\sqrt x=\frac{1+\sqrt 5}{2},$$ and so on.

Answer 2

$$ \sqrt{x} = \frac{1}{2} \pm \frac{\sqrt{5}}{2} $$

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