Given:
The temperature (in degrees Fahrenheit) at time t is given by
[tex]T(t)=66+144e^{-0.04t}[/tex]Required:
We need to find the time when the soup to cool to a temperature of 120°F.
Explanation:
Substitue T(t)=120 in the equation to find the value of t.
[tex]120=66+144e^{-0.04t}[/tex]Subtract 66 from both sides of the equation.
[tex]120-66=66+144e^{-0.04t}-66[/tex][tex]54=144e^{-0.04t}[/tex]Divide both sides by 144.
[tex]\frac{54}{144}=\frac{144e^{-0.04t}}{144}[/tex][tex]0.375=e^{-0.04t}[/tex]Take natural log on both sides of the equation,
[tex]In(0.375)=-0.04t[/tex][tex]-0.9809=-0.04t[/tex]Divide both sides by (-0.04).
[tex]\frac{-0.9809}{-0.04}=\frac{-0.04t}{-0.04}[/tex][tex]24.5225=t[/tex]Round of the nearest tenth of a minute.
[tex]24.5=t[/tex]We get t =24.5 minutes.
Final answer:
It takes 24.5 minutes for the soup to cool to a temperature of 120°F.
