Solve for $x$ in $\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$ I am trying to solve for $x$ but am very unsure of the steps. If someone could walk me through this I would really appreciate it.
$$\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$$

$$\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$$
Clearly  we can see that $4^{3/2}=\left(4^{\frac{1}{2}}\right)^3=2^3=8$. So 
$$\frac{\sqrt x-1}{x-1}>\frac{2^3}{2^4}$$
$$\frac{\sqrt x-1}{x-1}>\frac{1}{2}$$
$$\frac{2\sqrt x-2}{x-1}>1$$
$$\frac{2\sqrt x-2}{x-1}-1>0$$
$$\frac{2\sqrt x-2-x+1}{x-1}>0$$
$$\frac{2\sqrt x-x-1}{x-1}>0$$
Multiply both sides by $(-1)$ . Then ,
$$\frac{x-2\sqrt x+1}{x-1}<0$$
$$\frac{(\sqrt{x}-1)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}<0$$
Thus for $\sqrt{x} \neq1$ ,
$$\frac{\sqrt{x}-1}{\sqrt{x}+1}<0$$
Since $\sqrt{x}+1$ is always positive as $\sqrt{x}>0$ for all $x$ ,
$$\sqrt{x}-1<0$$
That is $$\sqrt{x}<1$$
$$0\leq x<1$$
