Question 1139494


The population {{{P}}} of a fish farm in {{{t}}} years is modeled by the equation
 
{{{P(t) = 1500/ (1 + 9e^(-0.8t)) }}}
 
To the nearest tenth, how many years will it take for the population to reach {{{900}}}? 


=> given {{{P(t) =900}}}


{{{900 = 1500/ (1 + 9e^(-0.8t)) }}} 

{{{1 + 9e^(-0.8t) = 1500/900 }}}

 {{{9e^(-0.8t) = 5/3-1 }}}

 {{{9e^(-0.8t) = 2/3}}}

 {{{e^(-0.8t) = (2/3)/9}}}

{{{ e^(-0.8t) = 2/27}}}...take natural log

{{{ln(e^(-0.8t) )= ln(2/27)}}} 

{{{(-0.8t) ln(e )= ln(2)-ln(3^3)}}}

{{{-0.8t *1= ln(2)-3ln(3)}}}

{{{-0.8t =-2.6}}}

{{{t=-2.6/(-0.8)}}}

{{{t}}}≈{{{3.3}}} years will it take for the population to reach {{{900}}}