we have the function
[tex]A(t)=A_0*e^{-0.0244t}[/tex]
Convert into an equivalent function of the form
[tex]y=a(b)^t[/tex]
so
[tex]A(t)=A_0e^{-0.0244t}=A_0(e^{-0.0244})^t=A_0(0.9759)^t[/tex]
therefore
the equivalent function is
[tex]A_0=A_0(0.9759)^t[/tex]
The rate of decay is equal to
b=1-r
b=0.9759
r=1-b
r=1-0.9759
r=0.0241
The answer Part a is -0.0241 (which is negative because is a decay rate)
Part b
For t=10 years
A_0=500 grams
substitute
[tex]\begin{gathered} A(t)=500*e^{-0.0244*(10)} \\ A(t)=392\text{ grams} \end{gathered}[/tex]
The answer part b is 392 grams
Part c
For A(t)=200 grams
Find out the value of t
substitute given values
[tex]\begin{gathered} 200=500e^{-0.0244(t)} \\ \frac{200}{500}=e^{-0.0244(t)} \end{gathered}[/tex]
Apply ln on both sides
[tex]\begin{gathered} ln\frac{200}{500}=lne^{-0.0244(t)} \\ \\ ln\frac{2}{5}=-0.0244t \end{gathered}[/tex]
t=37.6 years
The answer part c is 37.6 years
Part d
For A(t)=A_0/2
substitute
[tex]\begin{gathered} \frac{A_0}{2}=A_0e^{-0.0244(t)} \\ \frac{1}{2}=e^{-0.0244(t)} \end{gathered}[/tex]
Apply ln on both sides
[tex]ln\frac{1}{2}=lne^{-0.0244(t)}[/tex]
t=28.4 years
The answer part d is 28.4 years
