A pot of piping hot stew has been removed from the stove and left to cool. $\newline$ The relationship between the elapsed time, $m$ , in minutes, since the stew was removed from the stove, and the temperature of the stew, $T(m)$ , measured in ${ }^{\circ} C$ , is modeled by the following function. $\newline$ $T(m)=20+50 \cdot 10^{-0.04 m}$ $\newline$ How many minutes will it take for the stew to cool to a temperature of $30^{\circ} \mathrm{C}$ ? $\newline$ Round your answer, if necessary, to the nearest hundredth. $\newline$ $\square$ minutes

Full solution

Q. A pot of piping hot stew has been removed from the stove and left to cool.

\newline

The relationship between the elapsed time,

m

, in minutes, since the stew was removed from the stove, and the temperature of the stew,

T(m)

, measured in

{ }^{\circ} C

, is modeled by the following function.

\newline

T(m)=20+50 \cdot 10^{-0.04 m}

\newline

How many minutes will it take for the stew to cool to a temperature of

30^{\circ} \mathrm{C}

\newline

Round your answer, if necessary, to the nearest hundredth.

\newline

\square

minutes

Understand and Set up Equation: Understand the problem and set up the equation. $\newline$ We are given the temperature function $T(m) = 20 + 50 \times 10^{(-0.04m)}$ , where $T(m)$ is the temperature of the stew in degrees Celsius after $m$ minutes. We need to find the value of $m$ when $T(m) = 30\,\text{°C}$ . $\newline$ Set up the equation: $30 = 20 + 50 \times 10^{(-0.04m)}$ .
Isolate Exponential Term: Isolate the exponential part of the equation. $\newline$ Subtract $20$ from both sides of the equation to isolate the exponential term. $\newline$ $30 - 20 = 50 \times 10^{(-0.04m)}$ $\newline$ $10 = 50 \times 10^{(-0.04m)}$
Divide to Solve Exponential Term: Divide both sides by $50$ to solve for the exponential term. $\newline$ $\frac{10}{50} = 10^{(-0.04m)}$ $\newline$ $0.2 = 10^{(-0.04m)}$
Take Logarithm to Solve for $m$ : Take the logarithm of both sides to solve for $m$ . Use the property of logarithms that $\log(a^b) = b \cdot \log(a)$ . $\log(0.2) = \log(10^{-0.04m})$ $\log(0.2) = -0.04m \cdot \log(10)$ Since $\log(10)$ is $1$ , we can simplify this to: $\log(0.2) = -0.04m$
Solve for m: Solve for m. $\newline$ Divide both sides by $-0.04$ to isolate $m$ . $\newline$ $m = \frac{\log(0.2)}{-0.04}$ $\newline$ Calculate the value of $m$ using a calculator. $\newline$ $m \approx \frac{\log(0.2)}{-0.04} \approx 24.76$