Question 72953
If {{{A[o]}}} is the original amount, then {{{A[o]/2}}} is half of the original amount. So set {{{A[o]/2}}} equal to {{{A[o]e^(-0.058t)}}}
{{{A[o]/2=A[o]e^(-0.058t)}}}Now solve for t (remember e is a constant number)
{{{cross(A[o])/(2*cross(A[o]))=(cross(A[o])e^(-0.058t)/cross(A[o]))}}}Divide both sides by {{{A[o]}}}
{{{1/2=e^(-0.058t)}}}Now take the natural log of both sides (this undoes the natural base e)
{{{ln(1/2)=ln(e^(-0.058t))}}}
{{{(ln(1/2))/-0.058=cross(-0.058)/cross(-0.058)*t}}}Divide both sides by -0.058
{{{t=-ln(1/2)/0.058}}}
Since -ln(1/2)=0.693147 approximately
{{{-ln(1/2)/0.058=0.693147/0.058=11.9508}}}
So after 11.95 days or so the original amount remaining is half. Hope that helps.