Example227
According to recent research published by the Nature Publishing Group, the concentration of microplastic in the Great Pacific garbage patch (GPGP) is growing exponentially. In the year 1990 the concentration of micropastic was \(.4 \text{kg}/\text{km}^2\text{,}\) and in 2015 the concentration was \(1.2 \text{kg}/\text{km}^2\text{.}\)
Find a formula for the concentration in \(\text{kg}/\text{km}^2\) of microplastics in the GPGP \(t \) years after 1990.
How long will it take for the concentration to double?

The information provided in the problem is giving us two points on the graph of our exponential function. Starting with the exponential \(A(t)=a(b)^t\text{,}\) we begin trying to find the values of \(a\) and \(b \text{.}\) Since the function is to be in years since 1990, and the first value they gave us was for the year 1990, we immediately know that the value of \(a \) must be \(.4\text{.}\) Thus, our function now is read
\begin{equation*} A(t)=.4(b)^t. \end{equation*}Since 2015 is \(\alert{25}\) years after 1990, the second point they gave us now yields the equation
\begin{equation*} 1.2=.4(b)^{\alert{25}}. \end{equation*}Now, do not be fooled into thinking logarithms are needed for this equation. We only need logarithms when the unknown value is in the exponent, but the unknown here is the base. So, we simply divide each side by .4 and take the 25th root of each side. These steps look as follows: \begin{align*} 1.2\amp = .4(b)^{25}\\ \frac{1.2}{.4}\amp = b^{25}\\ \sqrt[25]{3}=b \end{align*} Then our formula is
\begin{equation*} A(t)=.4(\sqrt[25]{3})^t. \end{equation*} To identify the doubling time, we simply set our function equal to .8 as this is double the inital concentration of .4. Solving as we did in the previous section we have: \begin{align*} .8\amp =.4(\sqrt[25]{3})^t\ \ \text{now divide both sides by .4}\\ 2\amp = (\sqrt[25]{3})^t\ \ \text{apply log to each side}\\ \log(2)\amp = \log(\sqrt[25]{3})^t\ \ \text{use properties of logarithms to pull down t}\\ \log(2)\amp = t\cdot \log(\sqrt[25]{3})\ \ \text{divide each side by log(1.045)}\\ \frac{\log(2)}{\log(\sqrt[25]{3})}\amp = t. \end{align*} Thus, we see that \(t \approx 15.75\ \text{years}\text{.}\)