Section 1.9 : Exponential And Logarithm Equations
Exponential Growth/Decay. Many quantities in the world can be modeled (at least for a short time) by the exponential growth/decay equation.
\[Q = {Q_0}{{\bf{e}}^{k\,t}}\]If \(k\) is positive we will get exponential growth and if \(k\) is negative we will get exponential decay.
15. A population of bacteria initially has 250 present and in 5 days there will be 1600 bacteria present.
- Determine the exponential growth equation for this population.
- 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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We can start off here by acknowledging that we know,
\[Q\left( 0 \right) = 250\hspace{0.5in}{\rm{and}}\hspace{0.5in}Q\left( 5 \right) = 1600\]If we use the first condition in the equation we get,
\[250 = Q\left( 0 \right) = {Q_0}{{\bf{e}}^{k\left( 0 \right)}} = {Q_0}\hspace{0.5in} \to \hspace{0.5in}{Q_0} = 250\]We now know the first unknown in the equation. Plugging this as well as the second condition into the equation gives us,
\[1600 = Q\left( 5 \right) = 250{{\bf{e}}^{5k}}\]We can use techniques from earlier problems in this 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 earlier problems in this section how to do that. Here is the solution work for this part.
\[\begin{align*}\frac{{2000}}{{250}} & = {{\bf{e}}^{\frac{1}{5}\ln \left( {\frac{{32}}{5}} \right)t}}\\ \ln \left( 8 \right) & = \frac{1}{5}\ln \left( {\frac{{32}}{5}} \right)t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{5\ln \left( 8 \right)}}{{\ln \left( {{\textstyle{{32} \over 5}}} \right)}} = 5.6010}}\end{align*}\]It will take 5.601 days for the population to reach 2000.