Question 1158515
Assuming that the function describes the population in millions, we have {{{22e^(10k)=31}}}. Dividing both sides by 22, we have {{{e^(10k)=31/22}}}. Taking the natural logarithm of both sides, we get {{{ln(e^(10k))=ln(31/22)}}}. {{{10k=ln(31/22)}}} So {{{k=ln(31/22)/10}}}. Using a calculator to approximate, we have {{{k=0.034}}}.