Chapter 7 Exponential and Logarithmic Functions - 7.6 Solve Exponential and Logarithmic Equations - 7.6 Exercises - Problem Solving - Page 521: 54

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Given: $T=100\\T_0=200\\T_R=75\\r=0.054$ Apply Newton's Law of Cooling: $T=(T_0-T_R)e^{-rt}+T_R$ Substitute: $100=(200-75)e^{-(0.054)t}+75\\25=125e^{-(0.054)t}\\e^{0.054t}=\frac{125}{25}\\e^{0.054t}=5\\\ln e^{0.054t}=\ln 5\\0.054t=\ln 5\\t=\frac{\ln 5}{0.054}\approx29.8$