Question 1045259
{{{Q(t)}}} = {{{130(1-e^(-0.05t) ) +65}}}
{{{Q(t)}}} = {{{130(1-e^(-0.05t) ) +65}}}
a.
initial means {{{t}}}={{{0}}}
{{{Q(t=0)}}} = {{{130(1-e^(-0.05*0) ) +65}}} 
remember {{{x^0}}}={{{1}}}
{{{Q(t=0)}}} = {{{130(1-1 ) +65}}} 
{{{Q(t=0)}}} = {{{((130(0) ) +65)}}} 
{{{Q(t=0)}}} = {{{0+65}}} 
{{{Q(t=0)}}} = {{{65}}} 
b.
halfway through 20 week course is 10 at t=10
{{{Q(t)}}} = {{{130(1-e^(-0.05t) ) +65}}}


{{{Q(t=10)}}} = {{{130(1-e^(-0.05*10) ) +65}}} 
{{{Q(t=10)}}} = {{{130(1-e^(-0.5) ) +65}}} 
{{{Q(t=10)}}} = {{{130(1-0.60653) ) +65}}} 
{{{Q(t=10)}}} = {{{130(0.393) ) +65}}} 
{{{Q(t=10)}}} = {{{(51.15 ) +65}}} 
{{{Q(t=10)}}} = {{{116.15}}}
{{{Q(t=0)}}} = {{{116}}} 
 

c. 
means when {{{Q(t)}}}={{{100}}}
{{{Q(t)}}} = {{{130(1-e^(-0.05t) ) +65}}}
{{{Q(t)}}} = {{{130(1-e^(-0.05t) ) +65}}}
{{{100 }}}= {{{130(1-e^(-0.05*t) ) +65}}} 
{{{100 -65}}}= {{{130(1-e^(-0.05*t) ) }}} 
{{{35}}}={{{130(1-e^(-0.05*t) ) }}} 
{{{35}}}/{{{(130)}}}={{{(1-e^(-0.05*t) ) }}} 
{{{7}}}/{{{(26)}}}={{{(1-e^(-0.05*t) ) }}} 
{{{7}}}/{{{(26)}}}-{{{1}}}={{{(-e^(-0.05*t) ) }}} 
{{{7}}}/{{{(26)}}}-{{{26/26}}}={{{(-e^(-0.05*t) ) }}} 
{{{7-26}}}/{{{(26)}}}={{{(-e^(-0.05*t) ) }}} 

{{{-19}}}/{{{(26)}}}={{{(-e^(-0.05*t) ) }}} 

cancel - sign(minus)
{{{19}}}/{{{(26)}}}={{{(e^(-0.05*t) ) }}} 


{{{19}}}/{{{(26)}}}={{{(e^(-0.05*t))}}} 

take ln of both sides
ln({{{19}}}/{{{(26)}}}=ln{{{(e^(-0.05*t))}}}
{{{-0.1366}}}=ln{{{(e^(-0.05*t))}}}
{{{-0.1366}}}=(-0.05*t)*{{{(lne)}}}

remember {{{lne=1}}}
{{{(-0.1366)}}}={{{-0.05t*(1)}}}
{{{(-0.1366)}}}/{{{-0.05}}}={{{t}}}
{{{(6.27)}}}={{{t}}}
approxitely {{{t}}}={{{6}}} weeks

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