College Algebra (10th Edition)

$9.797$
We know that $x=2xe^{k\cdot1690}\\0.5=e^{1690k}\\1690k=\ln0.5\\k=\frac{\ln0.5}{1690}\approx-0.000410146$ Thus $N(50)=10e^{-0.000410146\cdot50}\approx9.797$