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. $\newline$ $C(t)=3+19e^{-0.045 t}$ $\newline$ 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. $\newline$ $\square$ 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+19e^{-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

\square

minutes

Set Temperature Function: Set the temperature function $C(t)$ equal to $16$ degrees Celsius to solve for $t$ . $\newline$ $C(t) = 3 + 19e^{(-0.045t)} = 16$
Subtract to Isolate Exponential Term: Subtract $3$ from both sides to isolate the exponential term. $\newline$ $19e^{(-0.045t)} = 13$
Divide to Solve Exponential Part: Divide both sides by $19$ to solve for the exponential part. $\newline$ $e^{(-0.045t)} = \frac{13}{19}$
Take Natural Logarithm: Take the natural logarithm ( $\ln$ ) of both sides to solve for the exponent. $\ln(e^{-0.045t}) = \ln(\frac{13}{19})$
Simplify Left Side: Use the property of logarithms that $\ln(e^x) = x$ to simplify the left side. $\newline$ $-0.045t = \ln(\frac{13}{19})$
Divide to Solve for t: Divide both sides by $-0.045$ to solve for $t$ . $\newline$ $t = \frac{\ln(\frac{13}{19})}{-0.045}$
Calculate Value of t: Calculate the value of t using a calculator. $\newline$ $t \approx \frac{\ln(\frac{13}{19})}{-0.045} \approx 20.1$