Question 282911
{{{P=P0*e^(kt)}}}
In one year, P will grow to 1.035*P0.
Use this to find k.
{{{1.035*P0=P0*e^(k)}}}
{{{e^k=1.035}}}
{{{k=ln(1.035)}}}
Now to find the time to double, let P=2*P0.
{{{2*P0=P0*e^(kt)}}}
{{{e^(kt)=2}}}
{{{kt=ln(2)}}}
{{{t=ln(2)/k=ln(2)/ln(1.035)}}}
That's the exact answer. 
Approximately, 
{{{t=20.15 }}}
About 20 years and 2 months.