<PRE><B>Question 1104542</B>
{{{y=1900b^x}}}


7 years ago, time 0


{{{1900b^0=1900}}}, understood from given


{{{1900b^7=900}}}
{{{b^7=9/19}}}
{{{log((b^7))=log((9))-log((19))}}}
{{{7*log((b))=log((9))-log((19))}}}

{{{log((b))=(log((9))-log((19)))/7}}}

.
.
{{{b=0.8988}}}


The model for population y is {{{highlight_green(y=1900(0.8988)^x)}}}.


A BETTER model would be now today, time is 0.  Choose then the 900 fish present NOW.
{{{highlight_green(highlight_green(y=900(0.8988)^x))}}} and x is years from today.
You just need to finish...</PRE>