The population of a particular type of bacteria doubles in number every 12 hours. a) If there...

Question:

The population of a particular type of bacteria doubles in number every 12 hours.

a) If there are 8 bacteria initially, write an equation that models the number of bacteria present after x = 12 hours.

b) How many bacteria will be present after 5 days?

Exponential Growth:

Exponential word problems with continuous growth or decay are a common real world application of mathematics. When given a starting quantity {eq}P {/eq} that is growing (or decaying) continuously at a rate {eq}r {/eq} over a given time {eq}t {/eq}, the ending amount {eq}A {/eq} can be found using:

{eq}A=Pe^{rt} {/eq}

Given the population of a particular type of bacteria doubles in number every {eq}12 {/eq} hours.

We can find the growth rate by using the equation:

{eq}A=Pe^{rt} {/eq}

Since the bacteria doubles every {eq}12 {/eq} hours, we will let {eq}A=2,\ P=1 {/eq}, and {eq}t=12 {/eq}:

{eq}2=1e^{r12}\\ 2=e^{r12} {/eq}

Take the natural log of both sides and solve for {eq}r {/eq}:

{eq}\ln 2=\ln e^{r12}\\ \ln 2=r12\ln e\\ \ln 2=r12\\ r=\frac{\ln 2}{12} {/eq}

a) If there are 8 bacteria initially, the equation that models the number of bacteria present after x = 12 hours is:

{eq}\boxed{\displaystyle{A=8e^{\left ( \frac{\ln 2}{12} t\right )}}} {/eq}

b) To find how many bacteria will be present after {eq}5 {/eq} days, we must keep in mind that {eq}5\ \text{days}=120\ \text{hours} {/eq}:

{eq}\begin{align*} A&=8e^{\left ( \frac{\ln 2}{12} \left (120 \right )\right )}\\ &=8e^{10\ln 2}\\ &=\boxed{8192\ \text{bacteria}} \end{align*} {/eq}