Growth of a bacterial culture

Statement

A culture initially contains $500$ bacteria and grows according to the model $N(t)=N_0\,e^{kt}$, with $t$ in hours. After $3$ hours there are $1800$ bacteria.

• (a) Find the constant $k$.
• (b) How many bacteria are there after $7$ hours?
• (c) After how long does the culture reach $10\,000$ bacteria?

Solution

(a) $1800=500\,e^{3k}\Rightarrow e^{3k}=3{.}6\Rightarrow k=\dfrac{\ln 3{.}6}{3}=\dfrac{1{.}28093}{3}\approx 0{.}4270\ \text{h}^{-1}$.

(b) $N(7)=500\,e^{0{.}4270\cdot 7}=500\,e^{2{.}98885}\approx 500\cdot 19{.}86\approx 9931$ cells.

(c) $10\,000=500\,e^{kt}\Rightarrow e^{kt}=20\Rightarrow t=\dfrac{\ln 20}{k}=\dfrac{2{.}99573}{0{.}4270}\approx 7{.}02$ hours.

$\boxed{k\approx 0{.}427\ \text{h}^{-1};\quad N(7)\approx 9931;\quad t\approx 7{.}02\ \text{h}}$