## Algebra and Trigonometry 10th Edition

We have $p(t)=\dfrac{1000}{1+9e^{-0.1656t}}$ Plug in $p(t)=500$ Therefore, $500=\dfrac{1000}{1+9e^{-0.1656}}$ or, $2= 1+9e^{-0.1656t}$ or, $e^{-0.1656t} =\dfrac{1}{9}$ or, $\ln e^{-0.1656t} =\ln \dfrac{1}{9}$ or, $t \approx 13$ months