Question 1114371
{{{F=P(1+i)^n}}}
{{{1000=500(1+0.08)^n}}}
{{{1.08^n=2}}}
{{{n=log(1.08,(2))}}}
{{{highlight(n=log((2))/log((1.08)))}}}
Find {{{n}}} then add it to 2018 to find the year.