180k views
5 votes
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= -----------

User Bzg
by
7.6k points

1 Answer

5 votes


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.

User StarPinkER
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories