A colony of bacteria is grown under ideal conditions in a laboratory so that the population...

Question:

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 2 hours, there are 4,800 bacteria. At the end of 4 hours, there are 19,000 bacteria. How many bacteria were present initially?

Exponential Growth:

We say that the quantity {eq}Q(t) {/eq} grows exponentially if {eq}Q(t)=Q(0)e^{kt} {/eq} for some positive constant {eq}k {/eq}. If a quantity grows exponentially, this means that the growth rate of the quantity is proportional to the current size of the quantity. Generally this means that nothing is restricting the growth of the quantity. For example, a population in a resource-rich environment may grow exponentially.

Let {eq}P(t) {/eq} be population of the bacteria colony after {eq}t {/eq} hours. Then we're given that {eq}P(2)=4800 {/eq} and {eq}P(4)=19000 {/eq}, and the value we're asked to compute is {eq}P(0) {/eq}.

Since the population of the colony grows exponentially, we know that {eq}P(t)=P(0)e^{kt} {/eq} for some constant {eq}t {/eq}. Using the known values for {eq}P(2) {/eq} and {eq}P(4) {/eq}, we have:

{eq}\begin{align*} 4800&=P(2)\\ &=P(0)e^{k\cdot 2}\\ &=P(0)e^{2k} \end{align*} {/eq}

and

{eq}\begin{align*} 19000&=P(4)\\ &=P(0)e^{k \cdot 4}\\ &=P(0)e^{4k} \, . \end{align*} {/eq}

Dividing the two equations, we give:

{eq}\begin{align*} \frac{P(0)e^{4k}}{P(0)e^{2k}}&=\frac{19000}{4800}\\ e^{2k}&=\frac{95}{24} \, . \end{align*} {/eq}

Substituting this back into the first equation, we have:

{eq}\begin{align*} P(0)e^{2k}&=4800\\ P(0)\left(\frac{95}{24}\right)&=4800\\ P(0)&=4800\left(\frac{24}{95}\right)\\ &=\frac{23040}{19}\\ &\approx 1212.6 \, . \end{align*} {/eq}

So the initial population of the bacteria is roughly {eq}\boxed{1213}\, {/eq}.