Give a formula for the concentration $C$ of lead in a pond $t$ years after being polluted by a factory, writing it in standard form $C=a b^{t}$. The initial concentration was $C_{0}$, and the lead levels drop by $30 \%$ every four years. State the values of $a$ and $b$, and say what they tell you in the context of the problem. NOTE: Round your answers to four decimal places. \[ C= \] \[ a= \] \[ b= \]

Step 1 ：Given that the initial concentration of lead was \(C_{0}\), we can say that \(a = C_{0}\).

Step 2 ：The problem also states that the lead levels drop by \(30\%\) every four years. This means that every four years, the concentration is \(70\%\) of what it was four years prior. Therefore, the decay factor \(b\) is \(0.70\) every four years.

Step 3 ：However, we want \(b\) to represent the decay factor per year, not per four years. To find this, we take the fourth root of \(0.70\) to find the yearly decay factor.

Step 4 ：So, let's calculate \(b\): \(b = 0.70^{1/4}\)

Step 5 ：Now, we can substitute \(a\) and \(b\) into the formula: \(C = C_{0} \cdot (0.70^{1/4})^{t}\)

Step 6 ：This is the formula for the concentration of lead in the pond \(t\) years after being polluted by a factory.

Step 7 ：Now, let's calculate the values of \(a\) and \(b\) to four decimal places: \(a = C_{0}\) and \(b = 0.70^{1/4} \approx 0.9123\)

Step 8 ：So, \(a = C_{0}\) tells us that the initial concentration of lead in the pond was \(C_{0}\), and \(b = 0.9123\) tells us that each year, the concentration of lead in the pond is approximately \(91.23\%\) of what it was the previous year.