Question 376325
a)


{{{A=500e^(-0.165t)}}} Start with the given equation.



{{{A=500e^(-0.165*5)}}} Plug in {{{t=5}}}



{{{A=500e^(-0.825)}}} Multiply -0.165 and 5 to get -0.825



{{{A=500(0.43823499246482)}}} Evaluate {{{e^(-0.825)}}} to approximately get {{{0.43823499246482}}}



{{{A=219.11749623241}}} Multiply



So approximately 219.117 mg (rounded to the nearest thousandth) will be left after 5 hrs.



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b)


{{{A=500e^(-0.165t)}}} Start with the given equation.



{{{100=500e^(-0.165t)}}} Plug in {{{A=100}}}



{{{100/500=e^(-0.165t)}}} Divide both sides by 500.



{{{0.2=e^(-0.165t)}}} Divide



{{{ln(0.2)=ln(e^(-0.165t))}}} Take the natural log of both sides.



{{{ln(0.2)=-0.165t*ln(e)}}} Pull down the exponent.



{{{ln(0.2)=-0.165t*1}}} Evaluate the natural log of 'e' to get 1.



{{{ln(0.2)=-0.165t}}} Multiply.



{{{(ln(0.2))/(-0.165)=t}}} Divide both sides by -0.165 to isolate 't'.



{{{9.75416916626728=t}}} Evaluate the left side with a calculator.



So the solution is approximately {{{t=9.754}}} (rounded to the nearest thousandth)



So in about 9.754 hours, there will be 100 mg left over.



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Jim