Answer:
8.6 grams of the substance remains after 30 years.
The amount of the substance will drop to below 20 grams in 14 years.
Step-by-step explanation:
The amount of the substance after t years is given by the following equation:
![A(t) = A(0)(1-r)^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20A%280%29%281-r%29%5E%7Bt%7D)
In which A(0) is the initial amount and r is the yearly decay rate.
A radioactive element decays at a rate of 5% annually. There are 40 grams of the substance present.
This means, respectively, that ![r = 0.05, A(0) = 40](https://tex.z-dn.net/?f=r%20%3D%200.05%2C%20A%280%29%20%3D%2040)
So
![A(t) = A(0)(1-r)^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20A%280%29%281-r%29%5E%7Bt%7D)
![A(t) = 40(0.95)^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%2040%280.95%29%5E%7Bt%7D)
How much of the substance remains after 30 years (to the nearest tenth)?
This is A(30).
![A(30) = 40(0.95)^{30} = 8.6](https://tex.z-dn.net/?f=A%2830%29%20%3D%2040%280.95%29%5E%7B30%7D%20%3D%208.6)
8.6 grams of the substance remains after 30 years
When will the amount of the substance drop to below 20 grams (to the nearest year)?
This is t when A(t) = 20. So
![A(t) = 40(0.95)^{t}](https://tex.z-dn.net/?f=A%28t%29%20%3D%2040%280.95%29%5E%7Bt%7D)
![20 = 40(0.95)^{t}](https://tex.z-dn.net/?f=20%20%3D%2040%280.95%29%5E%7Bt%7D)
![(0.95)^{t} = \frac{20}{40}](https://tex.z-dn.net/?f=%280.95%29%5E%7Bt%7D%20%3D%20%5Cfrac%7B20%7D%7B40%7D)
![(0.95)^{t} = 0.5](https://tex.z-dn.net/?f=%280.95%29%5E%7Bt%7D%20%3D%200.5)
![\log{(0.95)^{t}} = \log{0.5}](https://tex.z-dn.net/?f=%5Clog%7B%280.95%29%5E%7Bt%7D%7D%20%3D%20%5Clog%7B0.5%7D)
![t\log{0.95} = \log{0.5}](https://tex.z-dn.net/?f=t%5Clog%7B0.95%7D%20%3D%20%5Clog%7B0.5%7D)
![t = \frac{\log{0.5}}{\log{0.95}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clog%7B0.5%7D%7D%7B%5Clog%7B0.95%7D%7D)
![t = 13.51](https://tex.z-dn.net/?f=t%20%3D%2013.51)
Rounding to the nearest year
The amount of the substance will drop to below 20 grams in 14 years.