Question 143991
{{{N=P(1-e^(-0.15d))}}} Start with the given equation



{{{500=1000(1-e^(-0.15d))}}} Plug in {{{N=500}}} and {{{P=1000}}}



{{{500/1000=(1-e^(-0.15d))}}} Divide both sides by 1000



{{{1/2=1-e^(-0.15d)}}} Reduce



{{{1/2-1=-e^(-0.15d)}}} Subtract 1 from both sides



{{{-1/2=-e^(-0.15d)}}} Combine like terms



{{{(-1/2)/(-1)=e^(-0.15d)}}} Divide both sides by -1 (this will remove the negative sign from {{{-e}}}



{{{1/2=e^(-0.15d)}}} Divide



{{{ln(1/2)=ln(e^(-0.15d))}}} Take the natural log of both sides



{{{ln(1/2)=ln(e^(-0.15d))}}} Take the natural log of both sides



{{{ln(1/2)=-0.15d*ln(e)}}} Rewrite the expression using the identity  {{{ln(x^y)=y*ln(x))}}}



{{{ln(1/2)=-0.15d(1)}}} Take the natural log of "e" to get 1



{{{ln(1/2)=-0.15d}}} Multiply



{{{ln(1/2)/(-0.15)=d}}} Divide both sides by -0.15 to isolate d




So our answer is {{{d=-ln(1/2)/(0.15)}}}  which is approximately {{{d=4.62098}}}



So in a little more than 4 and half days, 500 students will have heard the rumor