In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Question

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Answer 1

Let the number of bacteria at a given times 't' be N
Then,

$\frac{d N}{d t} = k N$ where $k$ is the proportionality constant.

$\frac{d N}{N} = k d t$

$\int_{N_{0}}^{N} \frac{d N}{N} = \int_{0}^{t} k d t$

$l n (\frac{N}{N _{0}}) = k t$

$N = N_{0} e^{k t}$ ...(i)
If $N_{0} = 1 0^{5}$ and $t = 2 h o u r s$ and
$N = (1 + \frac{1 0}{1 0 0}) \times 1 0^{5}$

$= 1.1 \times 1 0^{5}$

Hence
$l n \frac{N}{N _{0}} = k t$

$l n (\frac{1 . 1 \times 1 0 ^{5}}{1 0 ^{5}}) = 2 k$

$l n (1.1) = 2 k$

$k = \frac{1}{2} l n (1.1) h o u r s^{- 1}$

$= l n (1.1) h o u r s^{- 1}$

Now
$l n (\frac{N}{N _{0}}) = k t$

N is now $2 \times 1 0^{5}$

Hence
$\frac{N}{N _{0}}$

$= \frac{2 \times 1 0 ^{5}}{1 0 ^{5}}$

$= 2$

Thus

$l n (\frac{N}{N _{0}}) = k t$ implies

$l n 2 = l n 1.1 t$

$l n 2 = \frac{l n ( 1 . 1 )}{2} t$

$2 l n 2 = l n (1.1) t$

$t = \frac{l n 4}{l n ( 1 . 1 )}$

$= 14.54$ hours

Answer 2

Let the number of bacteria at a given times 't' be N
Then,

\frac{d N}{d t} = k N

where

k

is the proportionality constant.

\frac{d N}{N} = k d t

\int_{N_{0}}^{N} \frac{d N}{N} = \int_{0}^{t} k d t

l n (\frac{N}{N _{0}}) = k t

N = N_{0} e^{k t}

...(i)
If

N_{0} = 1 0^{5}

and

t = 2 h o u r s

and

N = (1 + \frac{1 0}{1 0 0}) \times 1 0^{5}

= 1.1 \times 1 0^{5}

Hence

l n \frac{N}{N _{0}} = k t

l n (\frac{1 . 1 \times 1 0 ^{5}}{1 0 ^{5}}) = 2 k

l n (1.1) = 2 k

k = \frac{1}{2} l n (1.1) h o u r s^{- 1}

= l n (1.1) h o u r s^{- 1}

Now

l n (\frac{N}{N _{0}}) = k t

N is now

2 \times 1 0^{5}

Hence

\frac{N}{N _{0}}

= \frac{2 \times 1 0 ^{5}}{1 0 ^{5}}

= 2

Thus

l n (\frac{N}{N _{0}}) = k t

implies

l n 2 = l n 1.1 t

l n 2 = \frac{l n ( 1 . 1 )}{2} t

2 l n 2 = l n (1.1) t

t = \frac{l n 4}{l n ( 1 . 1 )}

= 14.54

hours