By how much should the distance from the source be increased to reduce the surface illuminance to 30 lumens? This is my problem and I have no idea how to solve it: 
The illuminance of a surface varies inversely with the square of its distance from the light source. 
If the illuminance of a surface is 120 lumens per square meter when its distance from a certain light source is 6 meters, by how many meters should the distance of the surface from the source be increased to reduce its illuminance to 30 lumens per square meter? 
Any help would be greatly appreciated.
 A: Let I represent illuminance, d represent distance from light source. Then, "The illuminance of a surface varies inversely with the square of its distance from the light source" implies the following equation: $$I=c/d^2$$ where $c$ is a constant. So, if I=120 when d=6, then $$120=c/(6^2)\implies c=120*36=4320$$ Now we know the constant $c$ so we can plug in 30 for $I$ and see what $d$ is: $$I=c/d^2\implies 30=4320/d^2\implies d^2=144\implies d=\pm12$$ But since distance cannot be negative, we use the positive root, namely $d=12$.
A: We want the illumination to decrease by a factor $\frac{30}{120}=\frac{1}{4}$.
So we want the distance to increase by a factor $\frac{1}{\sqrt{1/4}}=2$. 
The distance should be doubled, so it should be increased by $6$ metres. 
A: For a given distance $d$, you can get the luminance with
$$d^2=\frac{k}{l}$$
where $k$ is a constant. $l$ is your luminance, which varies inversely (hence the reciprocal of $l$) and with the square of the distance (hence $d^2$).
As such you have
$$(d_1)^2=\frac{k}{l_1}$$
and
$$(d_2)^2=\frac{k}{l_2}$$
Since these are proportional, you can combine these to make
$$\frac{(d_1)^2}{(d_2)^2}=\frac{kl_2}{kl_1}$$
$k$ can be eliminated to give
$$\frac{(d_1)^2}{(d_2)^2}=\frac{l_2}{l_1}$$
Plugging in your numbers,
$$\frac{6^2}{(d_2)^2}=\frac{30}{120}$$
$$\frac{36}{(d_2)^2}=0.25$$
$$\frac{36}{0.25}=(d_2)^2$$
$$12=d_2$$
And you're looking for the difference, so
$$d_2-d_1=12-6=6$$
