Question 1204990


{{{f(t) = 116(0.69)^t}}}  to  {{{f(t) = ae^(kt)}}}


{{{a* e^(k t) = 116* 0.69^t}}}....=>{{{a=116}}}

{{{116* e^(k t) = 116* 0.69^t}}}.....take log of both sides

{{{ln(116 *e^(k *t)) = ln(116* 0.69^t)}}}

{{{ln(116)+ln( e^(k *t)) = ln(116)+ln( 0.69^t)}}}

{{{(k* t)ln( e) = t*ln( 0.69)}}}......substituting {{{ln(e)=1}}}, then dividing by {{{t}}}, we have

{{{k = ln( 0.69)}}}


{{{a = 116}}}, {{{k = ln( 0.69)=-0.371}}}


{{{f(t) = 116e^(-0.371t)}}}