The solution of the differential equation ${\displaystyle {\frac {dT}{dt}}=k(T-T_{s})}$ is

${\displaystyle \displaystyle T-T_{s}=(T_{0}-T_{s})e^{kt}}$, where ${\displaystyle T}$ is the temperature of an object, ${\displaystyle T_{s}}$ is the temperature of its surroundings, and ${\displaystyle k}$ is some constant.

We are given that ${\displaystyle T_{s}=40,\,T_{0}=70,\,T(0.5)=60}$, where ${\displaystyle t}$ is in hours. Thus we begin by solving for ${\displaystyle k}$:

${\displaystyle {\begin{aligned}T(t)-T_{s}&=(T_{0}-T_{s})e^{kt}\\60-40&=(70-40)e^{k\cdot 0.5}\\e^{0.5k}&={\frac {2}{3}}\\{\frac {1}{2}}k&=\ln {\frac {2}{3}}\\k&=2\ln {\frac {2}{3}}=\ln \left({\frac {2}{3}}\right)^{2}=\ln {\frac {4}{9}}\\\\\therefore T(t)&=30e^{t\ln \left({\frac {4}{9}}\right)}+40=30e^{\ln \left[\left({\frac {4}{9}}\right)^{t}\right]}+40\\&=30\left({\frac {4}{9}}\right)^{t}+40\end{aligned}}}$

Finally, solving for ${\displaystyle t}$ when ${\displaystyle \displaystyle T(t)=50}$ yields:

${\displaystyle {\begin{aligned}50&=30\left({\frac {4}{9}}\right)^{t}+40\\{\frac {1}{3}}&=\left({\frac {4}{9}}\right)^{t}\\\ln {\frac {1}{3}}&=t\ln {\frac {4}{9}}\\t&={\frac {\ln {\frac {1}{3}}}{\ln {\frac {4}{9}}}}={\color {blue}{\frac {\ln 3}{\ln {\frac {9}{4}}}}\,\,\mathrm {hours} }\end{aligned}}}$