Question

A bacteria population grows by 10% every 2 years. Presently, the population is 80 000 bacteria. Find the population 12 years ago. (Can use log if needed but not “in”)

Answer 1

In this problem

we have an exponential growth function of the form

$y=a(1+r)^{\frac{t}{2}}$

where

r=10%=0.10

Let

t=0 ---------> 12 years ago

so

Presently -------> t=12 years, y=80,000 bacteria

substitute

$\begin{gathered} 80,000=a(1+0.10)^{\frac{12}{2}} \\ a=\frac{80,000}{1.10^6} \\ \\ a=45,158\text{ bacteria} \end{gathered}$

therefore

The population 12 years ago was 45,158 bacteria