[tex]\begin{gathered} \text{Given} \\ C(t)=85\mleft(\frac{1}{2}\mright)^{\frac{t}{6}} \end{gathered}[/tex]
[tex]\begin{gathered} \text{C. How long until there are only 20 mg remaining.} \\ \\ \text{Let }C(t)=20,\text{ then solve for }t \\ \\ C(t)=85\mleft(\frac{1}{2}\mright)^{\frac{t}{6}} \\ 20=85\mleft(\frac{1}{2}\mright)^{\frac{t}{6}} \\ \\ \text{Divide both sides by }85 \\ \frac{20}{85}=\frac{85\mleft(\frac{1}{2}\mright)^{\frac{t}{6}}}{85} \\ \frac{20}{85}=\mleft(\frac{1}{2}\mright)^{\frac{t}{6}} \\ \\ \text{Get the natural logarithm of both sides} \\ \mleft(\frac{1}{2}\mright)^{\frac{t}{6}}=\frac{20}{85} \\ \ln \mleft(\frac{1}{2}\mright)^{\frac{t}{6}}=\ln \frac{20}{85} \\ \frac{t}{6}\ln \mleft(\frac{1}{2}\mright)^{}=\ln \frac{20}{85} \\ \\ \text{Multiply both sides by }\frac{6}{\ln \frac{1}{2}} \\ \frac{6}{\ln\frac{1}{2}}\Big[\frac{t}{6}\ln (\frac{1}{2})^{}=\ln \frac{20}{85}\Big]\frac{6}{\ln\frac{1}{2}} \\ \frac{\cancel{6}}{\cancel{\ln \frac{1}{2}}}\Big{[}\frac{t}{\cancel{6}}\cancel{\ln (\frac{1}{2})}^{}=\ln \frac{20}{85}\Big{]}\frac{6}{\ln\frac{1}{2}} \\ \\ t=\frac{6\ln \frac{20}{85}}{\ln \frac{1}{2}} \\ t=12.52477705 \end{gathered}[/tex]
Therefore, it will take 12.52 hours
