[tex]P=250000e^{0.013t}\implies \stackrel{P}{450000}=250000e^{0.013t}\implies \cfrac{450000}{250000}=e^{0.013t} \\\\\\ \cfrac{9}{5}=e^{0.013t}\implies \log_e\left( \cfrac{9}{5} \right)=\log_e(e^{0.013t})\implies \ln\left( \cfrac{9}{5} \right)=0.013t \\\\\\ \cfrac{\ln\left( \frac{9}{5} \right)}{0.013}=t\implies 45.21\approx t\qquad \textit{about 45 years, 2 months and a half}[/tex]
