Question 374179
The equation to use is {{{x(t) = 500e^(-0.05t)}}}.
a.{{{  x(6) = 500e^(-0.05*6) = 500e^-0.3 = 370.409}}}grams.
b.  {{{250 = 500e^(-0.05t)}}}, {{{0.5 = e^(-0.05t)}}}, which gives {{{ln0.5 = -0.05t}}}, or {{{t = ln0.5/-0.05 = 13.86}}} hours.
c.  {{{200 = 500e^(-0.05t)}}} implies that {{{0.4 = e^(-0.05t)}}}, or {{{ln0.4 = -0.05t}}}, or {{{t = ln0.04/-0.05 = 18.33}}} hours.