Answer:
Step-by-step explanation:
Given the half like of a material to be 8.3 hours and the amount of iron-52 left after t hours is modeled by the equation [tex]A(t) = A_0 a^{t}[/tex], we can get A(t) as shown;
At t = 8.3 hours, A(8.3) = 1/2
Initially at t = 0; A(0) = 1
Substituting this values into the function we will have;
[tex]\frac{1}{2} = 1 * a^{8.3}\\\\Taking \ the \ log \ of\ both \ sides;\\\\log(\frac{1}{2} ) = log(a^{8.3} )\\\\log(\frac{1}{2} ) = 8.3 log(a)\\\\\fr-0.30103 = 8.3 log(a)\\\Dividing\ both\ sides\ by \ 8.3\\\\\frac{-0.30103}{8.3} = log(a)\\\\log(a) = - 0.03627\\\\a =10^{-0.03627} \\\\a = 0.919878 (to\ 6dp)[/tex]
