Problem Solving with Logs

In many cases - for long questions especially - you will have to deal with problems involving growth. Such problems often involve log manipulation and knowledge of how to handle $e$ and the natural log.

Example

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A scientist puts 100 bacteria into a test tube at the start of the experiment. Six hours later, he returns to the test tube and examines that there are 450 bacteria in the colony. The number of bacteria in the tube can be given by the equation :

y=Ae^{bt}

Where $b,A$ are constant, $t$ is the time (in hours).

Find the value of $A$ .
Find the value of $b$ (correct to two decimal places)

There are four unknowns altogether in this equation, which would make it infeasible to simply isolate and solve for $A$ . However, we are told that the scientist puts 100 bacteria at the start of the experiment. At the instance when the experiment begins, $t$ is 0, so let's substitute in for $t$ and for $y$ , the number of bacteria.

\begin{align*} 100&=Ae^{b(0)} & \\\\ &=Ae^{0} & \\\\ &=A(1) & \\\\ &=A \end{align*}

We solved for $A$ , the new equation looks like this :

y=100e^{bt}

To solve for $b$ , we could use the fact that the scientist observes 450 bacteria after 6 hours from when the experiment starts.

\begin{align*} 450&=100e^{6b} & \\\\ 4.5&=e^{6b} & \text{\footnotesize\textcolor{gray}{(\( \div100 \))}} \\\\ \ln{(4.5)}&=\ln{(e^{6b})} & \text{\footnotesize\textcolor{gray}{(\( \ln \))}} \\\\ &=6b\ln{(e)} & \text{\footnotesize\textcolor{gray}{(\( \log_a\left(x^q\right)=q\log_ax \))}} \\\\ &=6b & \text{\footnotesize\textcolor{gray}{(\( \ln{e}=1 \))}} \\\\ \frac{\ln(4.5)}{6}&=b & \text{\footnotesize\textcolor{gray}{(\( \div6 \))}} \\\\ :highlight[0.25]&=b & \end{align*}

Problem Solving with Logs Simplified Revision Notes for Leaving Cert Mathematics

Problem Solving with Logs

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