I need help with my homework question please it has two parts

ANSWER
[tex]\begin{gathered} (A)\text{ }1.36\text{ }hr \\ (B)\text{ }1.36\text{ }mg \end{gathered}[/tex]EXPLANATION
(A) To find the half-life of the drug, apply the exponential decay formula:
[tex]y=x(1-r)^t[/tex]where y = amount remaining
x = initial amount = 225 mg
r = decay rate = 40% = 0.4
t = time elapsed
Substituting the given values into the equation above:
[tex]\begin{gathered} y=225(1-0.4)^t \\ y=225(0.6)^t \end{gathered}[/tex]The half-life will occur when the amount of the drug is halved:
[tex]\begin{gathered} \frac{225}{2} \\ 112.5mg \end{gathered}[/tex]Substitute that for y and solve for t:
[tex]\begin{gathered} 112.5=225(0.6)^t \\ 0.5=0.6^t \end{gathered}[/tex]Convert the exponential equation into a logarithmic equation:
[tex]\begin{gathered} \log0.5=\log0.6^t \\ \Rightarrow\log0.5=t\log0.6 \\ \Rightarrow t=\frac{\log0.5}{\log0.6} \\ t=1.36\text{ }hr \end{gathered}[/tex]That is the half-life of the drug.
(B) To find the amount of the drug left after 10 hours, substitute 10 for t in the decay equation:
[tex]\begin{gathered} y=225*0.6^{10} \\ y=1.36\text{ }mg \end{gathered}[/tex]That is the answer.