# How many bacteria will be present after 5 hours if a culture of bacteria obeys the law of uninhibited growth where if 500 bacteria are present in the culture initially and there are 800 after 1 hour?

Jul 7, 2018

THe number of bacteria is $= 5243 \text{ bacteria}$

#### Explanation:

The growth of bacteria is according to the differential equation

$\frac{\mathrm{dN}}{\mathrm{dt}} = k N$

Integration of this equation yields

$\ln \left({N}_{2} / {N}_{1}\right) = k \left({t}_{2} - {t}_{1}\right)$

or

$\left({N}_{2} / {N}_{1}\right) = {e}^{k \left({t}_{2} - {t}_{1}\right)}$

The initial conditions are

${N}_{1} = 500$ at ${t}_{1} = 0$

Also,

${N}_{2} = 800$ at ${t}_{2} = 1 h$

Therefore,

$\ln \left(\frac{800}{500}\right) = k \left(1 - 0\right)$

Therefore,

$k = \ln \left(\frac{8}{5}\right)$

Therefore,

After ${t}_{3} = 5 h$, there are

$\ln \left({N}_{3} / {N}_{1}\right) = k \left({t}_{3} - {t}_{1}\right)$

$\ln \left({N}_{3} / 500\right) = \ln \left(\frac{8}{5}\right) \cdot 5 = 2.35$

${N}_{3} = 500 {e}^{2.35} = 5243 \text{ bacteria}$

Jul 7, 2018

$\textcolor{b l u e}{5242.88}$

#### Explanation:

We need to find an equation of the form:

$A \left(t\right) = A \left(0\right) {e}^{k t}$

Where:

$\boldsymbol{A \left(t\right)} =$ the amount after time t.

bb(A(0)= the amount at the start. i.e. t = 0.

$\boldsymbol{k} =$ the growth/decay factor.

$\boldsymbol{e} =$ Euler's number.

$\boldsymbol{t} =$ time, in this case hours.

We have:

$A \left(0\right) = 500$

$A \left(1\right) = 800$

Using these values:

$800 = 500 {e}^{k}$

We now solve for $\boldsymbol{k}$:

Divide by 500:

$\frac{8}{5} = {e}^{k}$

Taking natural logarithms of both sides:

$\ln \left(\frac{8}{5}\right) = k \ln \left(e\right)$

$\ln \left(e\right) = 1$ ( The logarithm of the base is always 1 ):

Hence:

$\ln \left(\frac{8}{5}\right) = k$

And:

$A \left(t\right) = 500 {e}^{t \ln \left(\frac{8}{5}\right)}$

This simplifies to:

$A \left(t\right) = 500 {\left(\frac{8}{5}\right)}^{t}$

Amount of bacteria after 5 hours:

$A \left(5\right) = 500 {\left(\frac{8}{5}\right)}^{5} = 5242.88$

Jul 7, 2018

If it obeys the law of uninhibited growth, growth occurs exponentially .

With $B$ as the number of bacteria:

• $B \left(t\right) = {B}_{o} {a}^{t} q \quad \left\{\begin{matrix}a \in \mathbb{R} \\ a > 1\end{matrix}\right.$

Use the given condition to find $a$:

$B \left(1\right) = 800 = 500 \cdot {a}^{1}$

$\implies a = \frac{8}{5}$

Then after 5 years:

$B \left(5\right) = 500 \cdot {\left(\frac{8}{5}\right)}^{5} = 5242.88$

Then round that to a whole number of bacteria :)