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A pot of piping hot stew has been removed from the stove and left to cool. 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 ^(@)C, is modeled by the following function. T(m)=20+50*10^(-0.04 m) How many minutes will it take for the stew to cool to a temperature of 30^(@)C ? Round your answer, if necessary, to the nearest hundredth. ◻ minutes

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

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Q. A pot of piping hot stew has been removed from the stove and left to cool.\newlineThe relationship between the elapsed time, m m , in minutes, since the stew was removed from the stove, and the temperature of the stew, T(m) T(m) , measured in ∘C { }^{\circ} C , is modeled by the following function.\newlineT(m)=20+50⋅10−0.04m T(m)=20+50 \cdot 10^{-0.04 m} \newlineHow many minutes will it take for the stew to cool to a temperature of 30∘C 30^{\circ} \mathrm{C} ?\newlineRound your answer, if necessary, to the nearest hundredth.\newline□ \square minutes
  1. Understand and Set up Equation: Understand the problem and set up the equation.\newlineWe are given the temperature function T(m)=20+50×10(−0.04m)T(m) = 20 + 50 \times 10^{(-0.04m)}, where T(m)T(m) is the temperature of the stew in degrees Celsius after mm minutes. We need to find the value of mm when T(m)=30 °CT(m) = 30\,\text{°C}.\newlineSet up the equation: 30=20+50×10(−0.04m)30 = 20 + 50 \times 10^{(-0.04m)}.
  2. Isolate Exponential Term: Isolate the exponential part of the equation.\newlineSubtract 2020 from both sides of the equation to isolate the exponential term.\newline30−20=50×10(−0.04m)30 - 20 = 50 \times 10^{(-0.04m)}\newline10=50×10(−0.04m)10 = 50 \times 10^{(-0.04m)}
  3. Divide to Solve Exponential Term: Divide both sides by 5050 to solve for the exponential term.\newline1050=10(−0.04m) \frac{10}{50} = 10^{(-0.04m)} \newline0.2=10(−0.04m) 0.2 = 10^{(-0.04m)}
  4. Take Logarithm to Solve for mm: Take the logarithm of both sides to solve for mm. Use the property of logarithms that log⁡(ab)=b⋅log⁡(a)\log(a^b) = b \cdot \log(a). log⁡(0.2)=log⁡(10−0.04m)\log(0.2) = \log(10^{-0.04m}) log⁡(0.2)=−0.04m⋅log⁡(10)\log(0.2) = -0.04m \cdot \log(10) Since log⁡(10)\log(10) is 11, we can simplify this to: log⁡(0.2)=−0.04m\log(0.2) = -0.04m
  5. Solve for m: Solve for m.\newlineDivide both sides by −0.04-0.04 to isolate mm.\newlinem=log⁡(0.2)−0.04m = \frac{\log(0.2)}{-0.04}\newlineCalculate the value of mm using a calculator.\newlinem≈log⁡(0.2)−0.04≈24.76m \approx \frac{\log(0.2)}{-0.04} \approx 24.76

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