## College Algebra (6th Edition)

$0.0039$
Let t be years after 2010. $A=A_{0}e^{kt }$, and we are given: $A_{0 }=82.3\quad$ (for t=0 in 2010) $A=70.5 \quad$when t=$40$, in 2050. From this we find k: $82.3=70.5e^{40k}\qquad/\div 70.5$ $\displaystyle \frac{82.3}{70.5}=e^{40k}\qquad$ ... apply ln( ) to both sides $\ln \displaystyle \frac{82.3}{70.5} =40k\qquad /\div 40$ $k= \displaystyle \frac{\ln \frac{82.3}{70.5}}{40}\approx$0.00386895994662$\approx 0.0039$