Question 73053
This equation models the decay of a radioactive element. As time goes on the element decays and we want to know how much of the original amount {{{A[o]}}} we should collect to have 40 grams after 3 days of decay. Since t represents the number of days, let t=3 and solve for {{{A[o]}}}
{{{40=A[o]e^(-0.18(3))}}}Set the decay model equal to 40 (after 3 days it decays to 40 grams. If you plug in t=3, you get 40)Note: e is a constant
{{{40/e^(-0.54)=(A[o]*cross(e^(-0.54)))/(cross(e^(-0.54)))}}}Divide both sides by {{{e^(-0.54)}}}
So 
{{{A[o]=40/e^(-0.54)}}}
{{{A[o]=40/0.582748}}}
{{{A[o]=40/0.582748}}}
{{{A[o]=68.64030
}}}Approximately
So about 68.6 grams should remain after 3 days

Check:
{{{A=(68.64030)e^(-0.18(t))}}}
If I let t=3 I should get A=40 (40 grams left over)
{{{A=(68.64030)e^(-0.18(3))}}}
{{{A=(68.64030)(0.582748)}}}
{{{A=39.9999975444
}}}Which is really close to 40, so this answer works

Hope this helps.