Henry places a bottle of water inside a cooler. As the water cools, its temperature $C(t)$ in degrees Celsius is given by the following function, where $t$ is the number of minutes since the bottle was placed in the cooler. $\newline$ $C(t)=3+19 e^{-0.045 t}$ $\newline$ Henry wants to drink the water when it reaches a temperature of $16$ degrees Celsius. How many minutes should he leave it in the cooler? $\newline$ Round your answer to the nearest tenth, and do not round any intermediate computations. $\newline$ minutes

Full solution

Q. Henry places a bottle of water inside a cooler. As the water cools, its temperature

C(t)

in degrees Celsius is given by the following function, where

t

is the number of minutes since the bottle was placed in the cooler.

\newline

C(t)=3+19 e^{-0.045 t}

\newline

Henry wants to drink the water when it reaches a temperature of

16

degrees Celsius. How many minutes should he leave it in the cooler?

\newline

Round your answer to the nearest tenth, and do not round any intermediate computations.

\newline

minutes

Set Temperature Function: First, we set the temperature function $C(t)$ equal to $16$ degrees Celsius to solve for $t$ . $\newline$ $16 = 3 + 19e^{-0.045t}$
Isolate Exponential Term: Subtract $3$ from both sides to isolate the exponential term. $\newline$ $16 - 3 = 19e^{-0.045t}$ $\newline$ $13 = 19e^{-0.045t}$
Solve for Exponent: Divide both sides by $19$ to solve for the exponent. $\newline$ $\frac{13}{19} = e^{-0.045t}$ $\newline$ $0.6842 = e^{-0.045t}$
Take Natural Logarithm: Take the natural logarithm ( $\ln$ ) of both sides to get rid of the exponential. $\newline$ $\ln(0.6842) = \ln(e^{-0.045t})$
Apply Logarithm Property: Use the property of logarithms that $\ln(e^x) = x$ . $\ln(0.6842) = -0.045t$
Solve for t: Divide by $-0.045$ to solve for $t$ . $\newline$ $t = \frac{\ln(0.6842)}{-0.045}$
Calculate Final Value: Calculate the value of $t$ using a calculator. $\newline$ $t \approx \ln(0.6842) / -0.045$ $\newline$ $t \approx 2.679 / -0.045$ $\newline$ $t \approx -59.533$