Question 1198693


{{{f(x) =1.8208*e^(0.3387x)}}}, where {{{x=0}}} corresponds to {{{1998}}}. 

In what year did/will the exports reach ${{{6.4}}} billion?

{{{f(x) =6.4}}}

{{{6.4=1.8208*e^(0.3387x)}}}..........take natural log of both sides

{{{ln(6.4)=ln(1.8208*e^(0.3387x))}}}

{{{ln(6.4)=ln(1.8208)+ln(e^(0.3387x))}}}

{{{ln(6.4)-ln(1.8208)=(0.3387x)ln(e)}}}...........{{{ln(e)=1}}}

{{{ln(6.4)-ln(1.8208)=0.3387x}}}

{{{x=(ln(6.4)-ln(1.8208))/0.3387}}}

{{{x=3.71131}}}years 

{{{x=3}}} years {{{8}}} months {{{16}}} days

if {{{x=0}}} corresponds to {{{1998}}}, then

{{{x=3.71131}}} corresponds to {{{1998+3}}}years {{{8}}} months {{{16}}} days

{{{2001}}}year {{{8}}} months {{{16}}} days

 the exports will  reach ${{{6.4}}} billion in {{{2001}}}