Answer
$8 \text{ feet}$
Work Step by Step
The variation model described by the problem is $
I=\dfrac{k}{x^2}
,$ where $I$ is the intensity of light and $x$ is the distance from the light source.
Substituting the known values of the variables results to
\begin{array}{l}\require{cancel}
80=\dfrac{k}{2^2}
\\\\
80=\dfrac{k}{4}
\\\\
4(80)=k
\\
k=320
.\end{array}
Hence, the equation of variation is
\begin{array}{l}\require{cancel}
I=\dfrac{320}{x^2}
.\end{array}
Using the variation equation above, then
\begin{array}{l}\require{cancel}
5=\dfrac{320}{x^2}
\\\\
5(x^2)=320
\\
5x^2=320
\\
x^2=\dfrac{320}{5}
\\\\
x^2=64
\\
x=\sqrt{64}
\\
x=8
.\end{array}
Hence, the distance of the light source, $x,$ is $
8 \text{ feet}
.$
