Answer
$x\approx1.22$ mm
Work Step by Step
We are given that the formula $\ln(\frac{I}{I_{0}})=-kx$ can be used to solve radiation problems (where x is the depth in millimeters, I is the intensity of radiation, $I_{0}$ is the initial intensity, and k is a constant measure dependent on the material).
In this case, we are given that $\frac{I}{I_{0}}=.02$ and $k=3.2$.
$\ln(.02)=-3.2x$
Divide both sides by -3.2.
$x=-\frac{\ln(.02)}{3.2}\approx1.22$ mm
