Question 1055302
{{{y=ae^bx}}}
So then,
{{{235000=ae^b(0)}}}
{{{a=235000}}}
So then 2005 is year 0, then 2016 will be year 11.
{{{308340=235000e^(11b)}}}
{{{e^(11b)=1.3124681}}}
{{{11b=0.27191}}}
{{{b=0.02472}}}
So then,
{{{y=235000e^(0.02472x)}}}
2025 would be year 20.
2020 would be year 15.
So for year 20,
{{{y=235000e^(0.02472(20))}}}
{{{y=385278}}}
Now do it for the year 15 in the same way.