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.
$$\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$$
 A: Note that $\frac{4^{3/2}}{2^4}=\frac12$ and the LHS can be simplified to $\frac1{\sqrt x+1}$. (The denominator is a difference of two squares: $x-1=(\sqrt x+1)(\sqrt x-1)$.)
$$\frac1{\sqrt x+1}>\frac12$$
Take reciprocals on both sides, which flips the inequality because both denominators are positive:
$$\sqrt x+1<2$$
$$\sqrt x<1$$
$$0\le x<1$$
A: First note that $\frac{4^{3/2}}{2^4} = \frac{1}{2}$. We then want to find the numbers $x$ such that
$$\frac{\sqrt{x}-1}{x-1} > \frac{1}{2}$$
To find these numbers, we will assume that equality holds, and then check each region (since the function on the LHS is continuous except at $x=1$, the only points when the inequality will go from true to false is when equality holds).
$$\frac{\sqrt{x}-1}{x-1} = \frac{1}{2}$$
$$2\sqrt{x}-2 = x-1$$
$$2\sqrt{x} = x+1$$
$$4x = x^2+2x+1$$
$$0 = x^2-2x+1$$
$$x-1 = 0$$
$$x = 1$$
We now must try some values of $x<1$ and some $x>1$. (At $x=1$ the value is not defined, but it is $\frac{1}{2}$ in the limit from both sides). One can easily check that at $x=\frac{1}{4}$, for instance, the inequality is satisfied, but at $x=4$, for instance, it is not. Thus, the inequality is satisfied only on the range $0\leq x<1$ (at $x<0$ the square root function is not real and thus cannot be compared in an inequality).
Edit: Here's a nicer answer:
Note that $x-1 = \left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)$ as a difference of squares. We then have that
$$\frac{1}{\sqrt{x}+1} > \frac{1}{2}$$
$$2 > \sqrt{x}+1$$
$$1 > \sqrt{x}$$
$$(0\leq)\ x<1$$
A: $$\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$$
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 $$\sqrt{x}<1$$
$$0\leq x<1$$
