# At a certain time, bacteria in a culture has 115,200 bacteria. Four hours later, the culture has 1,440,000 bacteria. How many bacteria will there be in the culture in 15 hours?

Feb 10, 2017

I got: $1 , 495 , 770 , 721$ BUT check my maths!

#### Explanation:

I would call the number of bacteria $y$ and write the exponential function:
$y = A {e}^{k t}$
with $t$ the time and $A \mathmr{and} k$ two constants.

With your numerical data we have:

At $t = {t}_{1}$

$115 , 200 = A {e}^{k {t}_{1}}$

At $t = {t}_{2} = {t}_{1} + 4$ (4 hours later):

$1 , 440 , 000 = A {e}^{k \left({t}_{1} + 4\right)}$
that can be written as:
$1 , 440 , 000 = A {e}^{k {t}_{1}} {e}^{4 k}$

we can substitute the first equation: $115 , 200 = A {e}^{k {t}_{1}}$ into the second to get:
$1 , 440 , 000 = \textcolor{red}{115 , 200} {e}^{4 k}$
solve it for $k$:
${e}^{4 k} = \frac{1 , 440 , 000}{115 , 200} = 12.5$
using natural logs on both sides we get:
$4 k = \ln \left(12.5\right)$
$k = \frac{1}{4} \ln \left(12.5\right) = 0.63143$

At $t = {t}_{3} = {t}_{1} + 15$ (15 hours later)

$y = A {e}^{k \left({t}_{1} + 15\right)}$
that can be written as:
$y = A {e}^{k {t}_{1}} {e}^{15 k}$
again we substitute the first equation and the value of $k$ found previously:
$y = 115 , 200 \cdot {e}^{15 \cdot 0.63143} = 1 , 495 , 770 , 721$

Feb 10, 2017

$1 , 495 , 770 , 721$ bacteria will be in $15$ hours.

#### Explanation:

Formula for exponential growth is
$y \left(t\right) = a \cdot {e}^{k t}$

Where y(t) = Number at time "t" , a = Number at start.
k = rate of growth (when >0) or decay (when <0) ,t = time

$1440000 = 115200 \cdot {e}^{k \cdot 4} \mathmr{and} {e}^{4 k} = \frac{1440000}{115200} = 12.5$ Taking log on both sides we get $k \cdot 4 = \ln \left(12.5\right) \mathmr{and} k = \ln \frac{12.5}{4} = 0.6314216$

After $15$ hrs bacteria will grow to $y \left(15\right) = 115200 \cdot {e}^{k \cdot 15} = 115200 \cdot {e}^{0.6314216 \cdot 15} = 1495770721$

$1 , 495 , 770 , 721$ bacteria will be in $15$ hours. [Ans]