Respuesta :
This is a type of radioactive decay problem. Given an initial amount of N0, the amount at t years is given by
[tex]N(t)=N_0e^{-\lambda t}[/tex]The information "27 years is the half-life of the substance" means that if we replace t by 27, we will get exactly half of the initial amount we had. That is
[tex]N_0e^{-\lambda\cdot27}=\frac{N_0}{2}[/tex]We can cancell out N0 on both sides, so we get
[tex]e^{-\lambda\cdot27}=\frac{1}{2}[/tex]using the properties of exponentials, we have the equivalent equation
[tex]\frac{1}{2}=\frac{1}{e^{\lambda\cdot27}}[/tex]which is also equivalent to
[tex]e^{\lambda\cdot27}=2[/tex]If we apply natural logarithm on both sides, we get
[tex]\ln (e^{\lambda\cdot27})=\lambda\cdot27=\ln (2)[/tex]Finally, we can divide both sides by 27, to get
[tex]\lambda=\frac{\ln (2)}{27}[/tex]So, the function that describes the amount of the subtances at time t is given as
[tex]N(t)=N_0e^{\frac{-\ln (2)\cdot t}{27}}[/tex]Now, we want to calculate the value of t, for which the amount we have at year t is exactly 66% of what we have at t=0. That is
[tex]\frac{N_0e^{-\frac{\ln (2)\cdot t}{27}}}{N_0}=0.66[/tex]We can cancell out N0 and, by equivalence by using the properties of exponents, we get
[tex]\frac{1}{e^{\frac{\ln(2)}{27}\cdot t}}=0.66[/tex]which is also equivalent to
[tex]\frac{1}{0.66}=e^{\frac{\ln (2)t}{27}}[/tex]if we apply the natural logarithm on both sides, we get
[tex]\ln (\frac{1}{0.66})=\ln (e^{\frac{\ln(2)\cdot t}{27}})=\ln (2)\cdot\frac{t}{27}[/tex]Finally, we want to solve for t. To do so, we can multiply by 27 and then divide by ln(2). So we get
[tex]t=\frac{27}{\ln(2)}\cdot\ln (\frac{1}{0.66})[/tex]with help of a calculator, we have that t is approximately 16.185. That is, about after 16 years, we will have 66% of the initial amount.
