Newton's law of cooling

Problem 6.395 · hard

1. T(t) = Tₐ + (T₀ − Tₐ)e^(−kt) solves dT/dt = −k(T − Tₐ).
   Reviewed
2. \[ \frac{d}{d t} \left(22 + 40 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{20}}\right) = - 2 e^{- \frac{t \ln{\left(\frac{3}{2} \right)}}{20}} \ln{\left(\frac{3}{2} \right)} \]
   The model satisfies the cooling law.✓ Proved
3. \[ \frac{146}{3} \]
   k = ln((T₀ − Tₐ)/(T₁ − Tₐ))/t₁ = log(3/2)/20 matches the reading at t = 20.✓ Proved
4. \[ \frac{80 \sqrt{6}}{9} + 22 \]
   T(30).✓ Proved
5. \[ \frac{20 \ln{\left(4 \right)}}{\ln{\left(\frac{3}{2} \right)}} = \ln{\left(2^{\frac{40}{\ln{\left(\frac{3}{2} \right)}}} \right)} \]
   Solve T(t) = target: e^(−kt) = (target − Tₐ)/(T₀ − Tₐ), so t = ln((T₀ − Tₐ)/(target − Tₐ))/k.✓ Proved