Question 926974
Let 9 years ago be {{{t=0}}}, then,
{{{1700=ae^(k(0))=a}}}
So today, {{{t=9}}}
{{{1000=1700e^(k(9))}}}
{{{e^(9k)=10/17}}}
{{{9k=ln(10/17)}}}
{{{k=(1/9)ln(10/17)}}}
So find {{{t}}} so that,
{{{1700e^(kt)=100}}}
{{{e^(kt)=1/17}}}
{{{kt=ln(1/17)}}}
{{{t=ln(1/17)/((1/9)ln(10/17))}}}
{{{t=9(ln(1)-ln(17))/(ln(10)-ln(17))}}}
{{{highlight(t=9(ln(17))/(ln(17)-ln(10)))}}}