Answer:
It will take 7 years ( approx )
Step-by-step explanation:
Given equation that shows the amount of the substance after t years,
![A=A_0 e^{kt}](https://tex.z-dn.net/?f=A%3DA_0%20e%5E%7Bkt%7D)
Where,
= Initial amount of the substance,
If the half life of the substance is 19 years,
Then if t = 19, amount of the substance =
,
i.e.
![\frac{A_0}{2}=A_0 e^{19k}](https://tex.z-dn.net/?f=%5Cfrac%7BA_0%7D%7B2%7D%3DA_0%20e%5E%7B19k%7D)
![\frac{1}{2} = e^{19k}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%3D%20e%5E%7B19k%7D)
![0.5 = e^{19k}](https://tex.z-dn.net/?f=0.5%20%3D%20e%5E%7B19k%7D)
Taking ln both sides,
![\ln(0.5) = \ln(e^{19k})](https://tex.z-dn.net/?f=%5Cln%280.5%29%20%3D%20%5Cln%28e%5E%7B19k%7D%29)
![\ln(0.5) = 19k](https://tex.z-dn.net/?f=%5Cln%280.5%29%20%3D%2019k)
![\implies k = \frac{\ln(0.5)}{19}\approx -0.03648](https://tex.z-dn.net/?f=%5Cimplies%20k%20%3D%20%5Cfrac%7B%5Cln%280.5%29%7D%7B19%7D%5Capprox%20-0.03648)
Now, if the substance to decay to 78% of its original amount,
Then ![A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0](https://tex.z-dn.net/?f=A%3D78%5C%25%20%5Ctext%7B%20of%20%7DA_0%20%3D%5Cfrac%7B78A_0%7D%7B100%7D%3D0.78%20A_0)
![0.78 A_0=A_0 e^{-0.03648t}](https://tex.z-dn.net/?f=0.78%20A_0%3DA_0%20e%5E%7B-0.03648t%7D)
![0.78 = e^{-0.03648t}](https://tex.z-dn.net/?f=0.78%20%3D%20e%5E%7B-0.03648t%7D)
Again taking ln both sides,
![\ln(0.78) = -0.03648t](https://tex.z-dn.net/?f=%5Cln%280.78%29%20%3D%20-0.03648t)
![-0.24846=-0.03648t](https://tex.z-dn.net/?f=-0.24846%3D-0.03648t)
![\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7](https://tex.z-dn.net/?f=%5Cimplies%20t%20%3D%20%5Cfrac%7B0.24846%7D%7B0.03648%7D%3D6.81085%5Capprox%207)
Hence, approximately the substance would be 78% of its initial value after 7 years.