Question 1129925

{{{P(t) = 1700/ (1 + (9^ (e-7t)))}}}


To the nearest tenth, how long will it take for the population to reach {{{900}}}?



{{{900= 1700/ (1 + (9^ (e-0.7t)))}}}


{{{900(1 + 9^ (e-0.7t))= 1700 }}}


{{{1 + 9^ (e-0.7t)= 1700/900 }}}


{{{9^ (e-0.7t)= 17/9-1 }}}


{{{9^ (e-0.7t)= 17/9-9/9 }}}


{{{(3^2)^ (e-0.7t)= 8/9 }}}


{{{3^ (2(e-0.7t))= 8/9 }}}.....take log of both sides


{{{log(3^(2(e-0.7t)))= log(8/9) }}}


{{{(2(e-0.7t))log(3)= log(2^3)-log(3^2) }}}


{{{(2(e-0.7t))log(3)= 3log(2)-2log(3) }}}


{{{e-0.7t= (3log(2))/(2log(3))-2log(3)/2log(3) }}}


{{{e-0.7t= (3log(2))/(2log(3))-1 }}}


{{{2.71828 - 0.7t= 0.9463946303571-1 }}}


{{{2.71828 - 0.7t=-0.0536053696428 }}}


{{{2.71828 +0.0536053696428= 0.7 t}}}


{{{t=2.7718853696428/0.7}}}


{{{t=3.9}}}