Question

Help please

A population of bacteria is growing according to the equation p(t)=1650e^0.11t Estimate when the population will exceed 2479.

t= -----------

Answer 1

`p(t)=1650e^(0.11t)\implies \stackrel{p(t)}{2479}=1650e^(0.11t)\implies \cfrac{2479}{1650}=e^(0.11t) \\\\\\ \log_e\left( \cfrac{2479}{1650} \right)=\log_e\left( e^(0.11t) \right)\implies \log_e\left( \cfrac{2479}{1650} \right)=0.11t \\\\\\ \ln\left( \cfrac{2479}{1650} \right)=0.11t\implies \cfrac{ ~~ \ln\left( (2479)/(1650) \right) ~~ }{0.11}=t\implies 3.7\approx t`

at that moment, the population will reach 2479, and right after that, it'll exceed it.