Answer:
[tex]A (t)=10e^{kt}[/tex]
Step-by-step explanation:
[tex]\frac{dA}{dt}\propto A(t)[/tex]
[tex]\frac{dA}{dt}=kA(t), where k=decay constant[/tex]
[tex]\frac{dA}{A}=k dt[/tex]
Taking the integral of both sides
[tex]\int{\frac{dA}{A}}=\int k dt[/tex]
[tex]ln A =kt+C\\[/tex], where C=constant of integration.
Taking the exponents of both sides
[tex]e^{ln A} =e^{kt+C}=e^{kt}Xe^{C}\\[/tex]
Since the exponential of a constant is still a constant
[tex]A (t)=Ce^{kt}[/tex]
At t=0, A(t)=10 g
[tex]10=C[/tex]
Therefore the initial value problem is given as:
[tex]A (t)=10e^{kt}[/tex]
