Question

a pie with a temperature of 140°f is taken out of the oven and placed on a windowsill to cool. its temperature as a function of t minutes is given by (t(t)=68e^{-0.0174t}+72). how long, to the closest minute, will it take for the pie to cool to 80°f? provide your answer below:

Explanation:

Step1: Set up the equation

Set $T(t)=80$, so $80 = 68e^{-0.0174t}+72$.

Step2: Isolate the exponential term

Subtract 72 from both sides: $80 - 72=68e^{-0.0174t}$, which gives $8 = 68e^{-0.0174t}$. Then divide both sides by 68: $\frac{8}{68}=e^{-0.0174t}$, or $\frac{2}{17}=e^{-0.0174t}$.

Step3: Take the natural - logarithm of both sides

$\ln(\frac{2}{17})=\ln(e^{-0.0174t})$. Since $\ln(e^{x}) = x$, we have $\ln(\frac{2}{17})=-0.0174t$.

Step4: Solve for t

We know that $\ln(\frac{2}{17})=\ln(2)-\ln(17)\approx0.693 - 2.833=-2.14$. Then $t=\frac{\ln(\frac{2}{17})}{-0.0174}=\frac{-2.14}{-0.0174}\approx123$.

Answer:

123