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### Section 6.5 : Applications

3. A population of bacteria initially has 250 present and in 5 days there will be 1600 bacteria present.

1. Determine the exponential growth equation for this population.
2. How long will it take for the population to grow from its initial population of 250 to a population of 2000?

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a Determine the exponential growth equation for this population. Show Solution

We can start off here by acknowledging that we know the initial number of bacteria is 250 and so $${Q_0} = 250$$. Therefore, the equation is then,

$Q\left( t \right) = 250{{\bf{e}}^{k\,t}}$

Now, we also know that $$Q\left( 5 \right) = 1600$$ and plugging this into the equation above gives,

$1600 = Q\left( 5 \right) = 250{{\bf{e}}^{5k}}$

We can use techniques from the Solve Logarithm Equations section to determine the value of $$k$$.

\begin{align*}1600 & = 250{{\bf{e}}^{5k}}\\ \frac{{1600}}{{250}} & = {{\bf{e}}^{5k}}\\ \ln \left( {\frac{{32}}{5}} \right) & = 5k\\ k & = \frac{1}{5}\ln \left( {\frac{{32}}{5}} \right) = 0.3712596\end{align*}

Depending upon your preferences we can use either the exact value or the decimal value. Note however that because $$k$$ is in the exponent of an exponential function we’ll need to use quite a few decimal places to avoid potentially large differences in the value that we’d get if we rounded off too much.

Putting all of this together the exponential growth equation for this population is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{Q = 250{{\bf{e}}^{\frac{1}{5}\ln \left( {\frac{{32}}{5}} \right)\,\,t}}}}$

b How long will it take for the population to grow from its initial population of 250 to a population of 2000? Show Solution

What we’re really being asked to do here is to solve the equation,

$2000 = Q\left( t \right) = 250{{\bf{e}}^{\frac{1}{5}\ln \left( {\frac{{32}}{5}} \right)\,\,t}}$

and we know from the Solve Logarithm Equations section how to do that. Here is the solution work for this part.

\begin{align*}\frac{{2000}}{{250}} & = {{\bf{e}}^{\frac{1}{5}\ln \left( {\frac{{32}}{5}} \right)\,\,t}}\\ \ln \left( 8 \right) & = \frac{1}{5}\ln \left( {\frac{{32}}{5}} \right)\,\,t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{5\ln \left( 8 \right)}}{{\ln \left( {\frac{{32}}{5}} \right)}} = 5.6010}}\end{align*}

It will take 5.601 days for the population to reach 2000.