Question 835283
{{{1=5(1/2)^(t/5730)}}}
{{{(1/2)^(t/5730)=1/5}}}
{{{t/5730=log(10,(1/5))/log(10,(1/2))}}}
{{{t=5730*(log(10,(1/5))/log(10,(1/2)))}}}
or approximately
{{{t=13305}}} years