We are given the initial function to be
[tex]\begin{gathered} A=A_0e^{-0.316t} \\ \text{where} \\ A_0=550mg \end{gathered}[/tex]
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Question 4
We are told to use the function above to estimate how long it will take for the amount in the body to be 175mg.
This simply translates to making A= 175 in the equation so that we will obtain
[tex]175=550e^{-0.316t}[/tex]
So we will make t the subject of the formula
To do so, we can follow the steps below
Step 1: Divide both sides by 550
[tex]\begin{gathered} \frac{175}{550}=\frac{550e^{-0.316t}}{550} \\ 0.318=e^{-0.316t} \end{gathered}[/tex]
Next, we will take the natural logarithm to both sides
[tex]\begin{gathered} ln(0.318)=\ln (e^{-0.316t}) \\ \ln 0.318=-0.316t \\ -0.316t=\ln 0.318 \end{gathered}[/tex]
Next, we will divide both sides by -0.316
[tex]\begin{gathered} t=\frac{\ln (0.318)}{-0.316} \\ t=3.6256 \end{gathered}[/tex]
Thus, the value of t = 3.6256 hours.
