Question 1057437
{{{P=P[0]e^(kt)}}}
{{{P=62e^(kt)}}}
So in 3 years,
{{{122=62e^(3k)}}}
{{{e^(3k)=122/62}}}
{{{e^(3k)=61/31}}}
{{{3k=ln(61/31)}}}
{{{k=ln(61/31)/3}}}
So then,
{{{P(6)=62e^((6/3)*ln(61/31))}}}
{{{P(6)=62e^(2*ln(61/31))}}}
{{{P(6)=122e^(ln(61/31))}}}
{{{P(6)=(122*61)/31}}}
{{{P(6)=7442/31}}}
{{{P(6)=240&2/31}}}
Since it must be an integer value,
{{{P(6)=240}}}