# A bacteria doubles its population in 8 hours. At this rate how many hours would it take the population of the bacteria to triple?

Nov 25, 2016

Time taken for 3 times the population to have grown is:

12 hours 40 minutes and say 47 seconds

#### Explanation:

Let the rate of growth be constant and of value y%
Let time in hours be $t$
Let count of bacteria at any time $t$ be ${b}_{t}$
So initial count of bacteria would be ${b}_{o}$

Given that count of bacteria after 8 hours (${b}_{8}$) is such that ${b}_{8} = 2 {b}_{o}$

Required to determine unknown time $x \text{ for } 3 {b}_{o} \to {t}_{x}$
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$\textcolor{b l u e}{\text{Building the model for initial condition}}$

The increase in bacteria after 1 hour is

${b}_{o} {\left(1 + \frac{y}{100}\right)}^{1}$

The increase in bacteria after 2 hour is

${b}_{o} {\left(1 + \frac{y}{100}\right)}^{2}$

The increase in bacteria after 2 hour is

${b}_{o} {\left(1 + \frac{y}{100}\right)}^{3}$

So after 8 hours we have:

$\textcolor{b l u e}{{b}_{o} {\left(1 + \frac{y}{100}\right)}^{8} = 2 {b}_{o}}$....................Equation(1)

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$\textcolor{b l u e}{\text{Determine the value of } y}$

Using equation(1) divide both sides by ${b}_{o}$

${\left(1 + \frac{y}{100}\right)}^{8} = 2$

Taking roots

$1 + \frac{y}{100} = \sqrt[8]{2}$

$\textcolor{b l u e}{y = 100 \left(\sqrt[8]{2} - 1\right)}$
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$\textcolor{b l u e}{\text{Determine the time it takes for } 3 {b}_{o}}$

Using Equation(1) we have:

${b}_{o} {\left(1 + \frac{y}{100}\right)}^{x} = 3 {b}_{o}$

$\implies {\left(1 + \frac{y}{100}\right)}^{x} = 3$

But $y = 100 \left(\sqrt[8]{2} - 1\right)$ giving

${\left(1 + \sqrt[8]{2} - 1\right)}^{x} = 3$

${\left(\sqrt[8]{2}\right)}^{x} = 3$

Taking logs ( I elect to use log to base 10 )

$x \log \left(\sqrt[8]{2}\right) = \log \left(3\right)$

But $\sqrt[8]{2} = {2}^{\frac{1}{8}}$ so we have $\log \left({2}^{\frac{1}{8}}\right) = \frac{1}{8} \log \left(2\right)$

$\frac{x}{8} \log \left(2\right) = \log \left(3\right)$

$x = \frac{8 \log \left(3\right)}{\log \left(2\right)}$

My calculator gives:

$x = 12.6797000058$

Is there could be a degree of error in the calculation lets say:

$x = 12.6797$ hours exactly

$x =$12 hours 40 minutes and say 47 seconds