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

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

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Answer 1 · 2016-11-25T18:01:19

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 $y %$ $y%$
Let time in hours be $t$ $t$
Let count of bacteria at any time $t$ $t$ be $b_{t}$ $b_t$
So initial count of bacteria would be $b_{o}$ $b_o$

Given that count of bacteria after 8 hours ( $b_{8}$ $b_8$ ) is such that $b_{8} = 2 b_{o}$ $b_8 =2b_o$

Required to determine unknown time $x for 3 b_{o} \to t_{x}$ $x" for "3b_o->t_x$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Building the model for initial condition$ $color(blue)("Building the model for initial condition")$

The increase in bacteria after 1 hour is

$b_{o} {(1 + \frac{y}{100})}^{1}$ $b_o(1+y/100)^1$

The increase in bacteria after 2 hour is

$b_{o} {(1 + \frac{y}{100})}^{2}$ $b_o(1+y/100)^2$

The increase in bacteria after 2 hour is

$b_{o} {(1 + \frac{y}{100})}^{3}$ $b_o(1+y/100)^3$

So after 8 hours we have:

$b_{o} {(1 + \frac{y}{100})}^{8} = 2 b_{o}$ $color(blue)(b_o(1+y/100)^8 = 2b_o)$ ....................Equation(1)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the value of y$ $color(blue)("Determine the value of "y)$

Using equation(1) divide both sides by $b_{o}$ $b_o$

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

Taking roots

$1 + \frac{y}{100} = \sqrt[8]{2}$ $1+y/100 = root(8)(2)$

$y = 100 (\sqrt[8]{2} - 1)$ $color(blue)(y=100(root(8)(2)-1))$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the time it takes for 3 b_{o}$ $color(blue)("Determine the time it takes for "3b_o)$

Using Equation(1) we have:

$b_{o} {(1 + \frac{y}{100})}^{x} = 3 b_{o}$ $b_o(1+y/100)^x = 3b_o$

$\Rightarrow {(1 + \frac{y}{100})}^{x} = 3$ $=> (1+y/100)^x=3$

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

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

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

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

$x log (\sqrt[8]{2}) = log (3)$ $xlog(root(8)(2))=log(3)$

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

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

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

My calculator gives:

$x = 12.6797000058$ $x=12.6797000058$

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

$x = 12.6797$ $x=12.6797$ hours exactly

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

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

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