<s>[INST] 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. 
 [/INST] 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
$$x_2 = \frac{p_1^2x_1}{p_2^2}$$
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