Logarithm Basic(4)Medium

This how I solve this problem:

The equation becomes: $\log(\frac{8x}{1+ \sqrt x})=2$

And because log are in base 10, we have:

$\frac{8x}{1+ \sqrt x}=100$

Rearrange:

$\sqrt x = \frac{8x}{100}-1$

$x=(\frac{2x}{25}-1)^2$ because $x\geq 0$

Expand:

$\frac{4x^2}{625}-\frac{29x}{25}+1=0$

There are two solutions: $x=\frac{\frac{29}{25} \pm \sqrt {(\frac{29}{25})^2-\frac{16}{625}}}{\frac{8}{625}} =25\frac{29 \pm \sqrt{29^2-16}}{8}$

They are approximately 0,866 and 180,38 but we need to care about the fact that $\sqrt x = \frac{2x}{25}-1$ and has to be positive, thus the only solution is the + one, and approximately $\boxed{180,38}$