Question 826467
{{{A=16.6*e^(0.547*t)}}}, for t years after year 2000, A in millions is the population at time t, 16.6 is population in millions at time t=0.


Part b:  When will A=23.3 ?
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Solve the equation for t, and then just substitute A=23.3.
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Take natural logarithm of both sides;
{{{ln(A)=ln(16.6e^(0.547t))}}}
{{{ln(A)=ln(16.6)+ln(e^(0.547t))}}}
{{{ln(A)=ln(16.6)+0.547t*ln(e)}}}
DO you know what to do from here?