Answer
$x=\left\{ \dfrac{1}{16},\dfrac{1}{3} \right\}$
Work Step by Step
Using $x^{-a}=\dfrac{1}{x^a}$, the given equation, $
x^{-2}-19x^{-1}+48=0
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{x^2}-\dfrac{19}{x}+48=0
.\end{array}
Multiplying both sides by the $LCD=
x^2
,$ then the solution to the given equation is
\begin{array}{l}\require{cancel}
1(1)-x(19)+x^2(48)=0
\\\\
1-19x+48x^2=0
\\\\
48x^2-19x+1=0
\\\\
(16x-1)(3x-1)=0
\\\\
x=\left\{ \dfrac{1}{16},\dfrac{1}{3} \right\}
.\end{array}
