Question 703916
Description of your model is {{{A=16.6e^(0.0547*t)}}}


a.    At year 2000, t=0, since t was described as years after 2000.
{{{A=16.6e^(0.0547*0)}}}
{{{A=16.6e^0=16.6*1=16.6}}}, million


b.    When will A=23.3 million?
You want to use the formula, solved for t.
If you really need the steps, they may be supplied, but I will tentatively omit them here; **See below.   you would begin by taking natural log of both sides of {{{A=16.6e^(0.0547*t)}}}.

Your formula for t will be or equivalent to:
{{{t=(ln(16.6)-ln(A))/(0.0547)}}}
Substitute your desired 23.3 for A and find t.


**From formula for A,
{{{ln(A)=ln(16.6*e^(0.0547t))}}}
{{{ln(A)=ln(16.6)+ln(e^(0.0547t))}}}
{{{ln(A)=ln(16.6)+0.0547*t*ln(e)}}}
{{{ln(A)=ln(16.6)+0.0547*t}}
{{{ln(A)-ln(16.6)=0.0547*t}}}
{{{t=(ln(A)-ln(16.6))/0.0547}}}