# Suppose an experiment begins with 5 bacteria, and the bacteria population triples every hour. What would be the population of the bacteria after 6 hours?

Aug 21, 2016

$= 3645$

#### Explanation:

$5 \times {\left(3\right)}^{6}$

$= 5 \times 729$

$= 3645$

Aug 21, 2016

$5 \times {3}^{6} = 3 , 645$

#### Explanation:

We could just write this out as $5 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

But this method would not be practical if we had to work it out for 24 hours, or for a week. If we can find a pattern or method, we will be able to work out the population for any time period.

Notice what we have done:

after 1 hour has elapsed, multiply by 3 once. $\times 3$
after 2 hours have elapsed, multiply by 3 twice. $\times {3}^{2}$
after 3 hour have elapsed, multiply by 3 thrice. $\text{ } {3}^{3}$
After 4 hours have elapsed, multiply by 3 , 4 times, or ${3}^{4}$

Now we can see that there is a pattern emerging.

Population = $5 \times {3}^{\text{number of hours}}$

=$5 \times {3}^{6} = 3 , 645$

If we treat this as a GP, note that we are actually looking for the value of the 7th term, because we started with 5, but the growth in the population is only seen AFTER 1 hour, from the 2nd term.

.

Aug 21, 2016

The population of Bacteria after $6$ hours$= 3645$.

#### Explanation:

In the beginning of the experiment, no. of bacteria$= 5$

As is given, after $1$ hour, the population$= {3}^{1} \cdot 5$.

After $2$ hours, the pop.$= 3 \left({3}^{1} \cdot 5\right) = {3}^{2} \cdot 5$

After $3$ hours, the pop.$= 3 \left({3}^{2} \cdot 5\right) = {3}^{3} \cdot 5$.

Clearly, after $6$ hours, the pop.$= {3}^{6} \cdot 5 = 3645$.

In general, the Population after $h$ hours$= 5 \cdot {3}^{h}$.

Enjoy Maths.!