Basic Math, exponents and algebra 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. 
 A: 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
$$
x_2=\frac{p_1^2}{p_2^2}x_1
$$
A: 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
$$x_2 = \frac{p_1^2x_1}{p_2^2}$$
A: 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?
A: 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*}
