How to solve for x in log(8x)-log(1+x^1/2)=2

In this case Melody, it might be better to let y = √x  (a positive number) when you get to 0.08x - 1 = √x  so that you have  0.08y² - 1 = y.  Then when you solve for y, if you get a negative number you can discard it immediately:

$${\mathtt{0.08}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{y}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,-\,}}{\mathtt{25}}\right)}{{\mathtt{4}}}}\\
{\mathtt{y}} = {\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{25}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.930\: \!703\: \!308\: \!172\: \!535\: \!8}}\\
{\mathtt{y}} = {\mathtt{13.430\: \!703\: \!308\: \!172\: \!535\: \!8}}\\
\end{array} \right\}$$

Taking only the positive value we then have (approximately)

$${\mathtt{x}} = {{\mathtt{13.430\: \!7}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{x}} = {\mathtt{180.383\: \!702\: \!49}}$$

.

$$\\log(8x)-log(1+x^{1/2})=2\\\\
log\frac{8x}{1+x^{0.5}}=2\\\\
10^{log\frac{8x}{1+x^{0.5}}}=10^2\\\\
\frac{8x}{1+x^{0.5}}=100\\\\
8x=100(1+x^{0.5})\\\\
0.08x=1+\sqrt{x}\\\\
0.08x-1=\sqrt{x}\\\\
0.0064x^2-0.16x+1=x\\\\
0.0064x^2-1.16x+1=0\\\\$$

x is approx  0.866  or  180.384  but these must be checked. :)

$${\mathtt{0.006\: \!4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1.16}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{125}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,-\,}}{\mathtt{725}}\right)}{{\mathtt{8}}}}\\
{\mathtt{x}} = {\frac{\left({\mathtt{125}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{33}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{725}}\right)}{{\mathtt{8}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.866\: \!208\: \!647\: \!843\: \!302\: \!2}}\\
{\mathtt{x}} = {\mathtt{180.383\: \!791\: \!352\: \!156\: \!697\: \!8}}\\
\end{array} \right\}$$

log(8x)-log(1+x^1/2)=2

check

$${log}_{10}\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{0.866\: \!208\: \!643}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{0.866\: \!208\: \!643}}}^{{\mathtt{0.5}}}\right) = {\mathtt{0.554\: \!996\: \!959\: \!640\: \!612\: \!2}}$$

$${log}_{10}\left({\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{180.383\: \!791\: \!352}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{180.383\: \!791\: \!352}}}^{{\mathtt{0.5}}}\right) = {\mathtt{1.999\: \!999\: \!999\: \!999\: \!798\: \!4}}$$

$$So\;\; x\approx 180.184$$

Yes that is a much better idea - thanks Alan.