In 2009, there were 1570 bears in a wildlife refuge. In 2010, the population had increased toapproximately 1884 bears. If this trend continues and the bear population is increasing exponentially,how many bears will there be in 2018?

Answer:[tex]8,101\ bears[/tex]Step-by-step explanation:we know thatIn this problem we have a exponential function of the form[tex]y=a(b)^{x}[/tex]wherex ----> is the number of years since 2009y ----> is the population of bearsa ----> is the initial value b ---> is the basestep 1Find the value of aFor x=0 (year 2009)y=1,570 bearssubstitute[tex]1.570=a(b)^{0}[/tex][tex]a=1.570\ bears[/tex]so[tex]y=1.570(b)^{x}[/tex]step 2Find the value of bFor x=1 (year 2010)y=1,884 bearssubstitute[tex]1,884=1.570(b)^{1}[/tex][tex]b=1,884/1.570[/tex][tex]b=1.2[/tex]The exponential function is equal to[tex]y=1.570(1.2)^{x}[/tex]step 3How many bears will there be in 2018?2018-2009=9 yearssoFor x=9 yearssubstitute in the equation[tex]y=1.570(1.2)^{9}[/tex][tex]y=8,101\ bears[/tex]