Question

use the exponential decay model, a = a₀eᵏᵗ, to solve the following. the half - life of a certain substance is 25 years. how long will it take for a sample of this substance to decay to 72% of its original amount? it will take approximately for the sample of the substance to decay to 72% of its original amount. (round the final answer to one decimal place as needed. round all intermediate values to four decimal places as needed.)

Explanation:

Step1: Find the value of k using half - life

When the substance reaches half - life, $A=\frac{A_0}{2}$ and $t = 25$. Substitute into $A = A_0e^{kt}$:
$\frac{A_0}{2}=A_0e^{25k}$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $\frac{1}{2}=e^{25k}$. Take the natural logarithm of both sides: $\ln(\frac{1}{2})=\ln(e^{25k})$. Since $\ln(e^{x}) = x$, we have $\ln(\frac{1}{2})=25k$. Then $k=\frac{\ln(\frac{1}{2})}{25}\approx\frac{- 0.6931}{25}=-0.0277$.

Step2: Find the time t when $A = 0.72A_0$

Substitute $A = 0.72A_0$ and $k=-0.0277$ into $A = A_0e^{kt}$:
$0.72A_0=A_0e^{-0.0277t}$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $0.72=e^{-0.0277t}$. Take the natural logarithm of both sides: $\ln(0.72)=\ln(e^{-0.0277t})$. Since $\ln(e^{x}) = x$, we have $\ln(0.72)=-0.0277t$. Then $t=\frac{\ln(0.72)}{-0.0277}\approx\frac{-0.3283}{-0.0277}\approx11.9$.

Answer:

$11.9$