Question 1023393
For continuously compounded interest, we have the equation
{{{A(t) = A[0]e^(rt)}}}
so we plug in to find t...
{{{10000 = 5000e^(.022t)}}}
{{{2 = e^(.022t)}}}
{{{ln(2) = .022t}}}
{{{t = ln(2)/.022}}} and
t = 31.5 years