Answer
$4m$
Work Step by Step
Step 1. Write the general formula of light intensity as $I(r)=\frac{k}{r^2}$, where $r$ is the distance from the light source and $k$ is a constant.
Step 2. Draw a diagram of the configuration as shown (please note that we can only solve a one-dimensional case for this problem with points in between the two lights). The stronger light source is at $(0,0)$ and the weaker one is at $(6,0)$. The relation between their $k$ factors is $k_1=8k_2=8k$
Step 3. For a point $P(x,0), 0\lt\ x\lt6$, the total light intensity at this point is given by $I=\frac{k_1}{r_1^2}+\frac{k_2}{r_2^2}$, where $r_1=x$ and $r_2=6-x$. We need to find the minimum in $I$ at a certain $x$.
Step 4. Combine the above results, $I=\frac{8k}{x^2}+\frac{k}{(6-x)^2}$. Take its derivative to get $I'=-\frac{16k}{x^3}+\frac{2k}{(6-x)^3}$. Letting $I'=0$, we have $8(6-x)^3=x^3$, which gives $x=4m$
Step 5. Check $I''=\frac{48k}{x^4$}+\frac{6k}{(6-x)^4}\gt0$ and the region is concave up with a local minimum.
