# A radioactive substance decays at a rate of 30% per year. Presently there are 400 grams of the substance. How long before there are 10 grams?

May 28, 2016

$\ln \frac{40}{0.3}$ years = 12.296 years = 12 years and 108 days, nearly.

#### Explanation:

If x gm is present at time t years,

$x ' = - 0.3 x$ gm/year. Integrating,

$\int \frac{1}{x} \mathrm{dx} = - 0.3 \int \mathrm{dt}$.

So, $\ln x = - 0.3 t + A$. Inversely,

$x = {c}^{- 0.3 t + A} = {e}^{A} {e}^{- 0.3 t} = C {e}^{- 0.3 t}$

Initially, t = 0 and x = 400. So, C=400, and now,

$x = 400 {e}^{- 0.3 t}$

The time t years for decay, from 400 gm to 10 gm, is given by

$10 = 400 {e}^{- 0 , 3 t}$. So, ${e}^{0.3 t} = 40$, and inversely,

$0.3 t = \ln 40$.

So, t=ln 40/0.3=12.296 years= 12 years and 108 days, nearly.