Given,
[tex]T=Ae^{-kt}+C[/tex]
C = 73°
A = 174°
T = 131°
k = 0.0688919
find t,
Steps,
#1 Replace
[tex]131=174*e^{-0.0688919t}+73[/tex]
#2. Isolate e^-kt
[tex]\frac{131-73}{174}=e^{-0.0688919t}[/tex]
#3. Simplify the left hand side
[tex]\frac{1}{3}=e^{-0.0688919t}[/tex]
#4. Apply exponent rules
[tex]\begin{gathered} ln(\frac{1}{3})=ln(e^{-0.0688919t}) \\ ln(\frac{1}{3})=-0.0688919t \end{gathered}[/tex]
#5. Solve for t
[tex]t=\frac{ln(\frac{1}{3})}{-0.0688919}=\frac{ln(3)}{0.0688919}\approx15.94690...[/tex]
#6. Round to two decimal places
[tex]t\approx15.95[/tex]
Answer: 15.95
