I have the equation $$\frac{x_1^{-\frac{1}{2}}}{{x_2^{-\frac{1}{2}}}} = p_l/p_2$$
How do I get $x_2$ on its own?
I have $$x_2^{-\frac{1}{2}} = \frac{p_2(x_1^{-\frac{1}{2}})}{p_1}$$ And if you have a useful link that reviews this info, it would be highly appreciated.
Square both sides: $$ \frac{x_1^{-1}}{x_2^{-1}}=\frac{p_1^2}{p_2^2} $$ Rewrite the left hand side: $$ \frac{x_1^{-1}}{x_2^{-1}}= \frac{1/x_1}{1/x_2}=\frac{1}{x_1}\left(\frac{1}{x_2}\right)^{-1}= \frac{1}{x_1}x_2=\frac{x_2}{x_1} $$ Thus your equality is $$ \frac{x_2}{x_1}=\frac{p_1^2}{p_2^2} $$ or $$ x_2=\frac{p_1^2}{p_2^2}x_1 $$
By definition, if $x \neq 0$, then
$$x^{-n} = \frac{1}{x^n}$$
Thus,
$$x_1^{-\frac{1}{2}} = \frac{1}{x_1^{\frac{1}{2}}}$$
If we make the substitution $n = -m$ in the equation
$$x^{-n} = \frac{1}{x^n}$$
we obtain
$$x^{m} = \frac{1}{x^{-m}}$$
Hence,
$$\frac{1}{x_2^{-\frac{1}{2}}} = x_2^{\frac{1}{2}}$$
Therefore, we can rewrite the equation
$$\frac{x_1^{-\frac{1}{2}}}{x_2^{-\frac{1}{2}}} = \frac{p_1}{p_2}$$
in the form
$$\frac{x_2^{\frac{1}{2}}}{x_1^{\frac{1}{2}}} = \frac{p_1}{p_2}$$
which we can solve for $x_2$ by multiplying both sides by $x_1^{\frac{1}{2}}$ to obtain
$$x_2^{\frac{1}{2}} = \frac{p_1x_1^{\frac{1}{2}}}{p_2}$$
then squaring both sides to obtain
$$x_2 = \frac{p_1^2x_1}{p_2^2}$$
You're almost there: from $x_2^{-1/2} = \frac{p_2x_1^{-1/2}}{p_1}$, take the reciprocal of both sides and then square both sides: \begin{align*} x_2^{-1/2} &= \frac{p_2x_1^{-1/2}}{p_1} \\ x_2^{1/2} &= \frac{p_1}{p_2x_1^{-1/2}} \\ x_2 &= \frac{p_1^2}{p_2^2x_1^{-1}} = \frac{x_1p_1^2}{p_2^2}. \end{align*}
Hint
If you have $$u^{-\frac{1}{2}}=v$$ square both sides to get $$u^{-1}=v^2$$ Now just re-arrange the variables: $$u^{-1}v^{-2}=1$$ $$v^{-2}=u$$ Can you go from here?