Answer
$0.0039$
Work Step by Step
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$