a) The initial temperature of the liquid is simply the temperature of the liquid at time m = 0
Thus, we have the temperature to be:
[tex]\begin{gathered} T(m)=101e^{-0.03m}+67 \\ \Rightarrow T(0)=101e^{-0.03\times0}+67 \\ \Rightarrow T(0)=101e^0+67 \\ \Rightarrow T(0)=101(1)^{}+67 \\ \Rightarrow T(0)=101^{}+67 \\ \Rightarrow T(0)=168^oF \end{gathered}[/tex]The initial temperature of the liquid is 168 degrees Fahrenheit
b) The time when the temperature of the liquid will reach 100 degrees Fahrenheit is obtained as follows:
[tex]\begin{gathered} T(m)=101e^{-0.03m}+67 \\ \Rightarrow100=101e^{-0.03m}+67 \\ \Rightarrow100-67=101e^{-0.03m} \\ \Rightarrow33=101e^{-0.03m} \\ \Rightarrow\frac{33}{101}=e^{-0.03m} \\ \Rightarrow e^{\mleft\{-0.03m\mright\}}=\frac{33}{101} \end{gathered}[/tex]Now, take the natural logarithm of both sides, as follows:
[tex]\begin{gathered} e^{\mleft\{-0.03m\mright\}}=\frac{33}{101} \\ \Rightarrow\log _ee^{\{-0.03m\}}=\log _e(\frac{33}{101}) \\ \Rightarrow(-0.03m)\log _ee=\log _e(\frac{33}{101}) \\ \Rightarrow(-0.03m)(^{}1)=\log _e(\frac{33}{101}) \\ \Rightarrow-0.03m=\log _e(\frac{33}{101}) \\ \Rightarrow m=\frac{1}{-0.03}\times\log _e(\frac{33}{101}) \\ \Rightarrow m=\frac{1}{-0.03}\times-1.1186=\frac{-1.1186}{-0.03}=37.39 \\ \Rightarrow m=37.4\text{ minutes (to the nearest tenth of a minute)} \end{gathered}[/tex]The time when the temperature of the liquid will reach 100 degrees Fahrenheit is 37.4 minutes
