Question

question 7 (multiple choice worth 5 points)
(exponential functions mc)
a video game tournament begins with 243 competitors. after each round, one - third of the competitors remain. write the function that models the change between the number of rounds
x, and the number of competitors remaining, r(x).
r(x)=243cdot(\frac{1}{3})^{x}
r(x)=3cdot(\frac{1}{243})^{x}
r(x)=243cdot3^{x}
r(x)=3cdot243^{x}

Explanation:

Step1: Recall the exponential decay formula

The general form of an exponential decay function is \(y = a\cdot b^{x}\), where \(a\) is the initial amount and \(b\) is the base (\(0 < b<1\) for decay). Here, the initial number of competitors \(a = 243\), and the base \(b=\frac{1}{3}\) (since one - third of the competitors remain each round, meaning the number of competitors is multiplied by \(\frac{1}{3}\) each round).

Step2: Substitute \(a\) and \(b\) into the formula

Substituting \(a = 243\) and \(b=\frac{1}{3}\) into the formula \(r(x)=a\cdot b^{x}\), we get \(r(x)=243\cdot(\frac{1}{3})^{x}\). We can also rewrite \(243\) as \(3^{5}\), and \((\frac{1}{3})^{x}=3^{-x}\). Then \(r(x)=3^{5}\cdot3^{-x}=3^{5 - x}\). Another way is to note that \(243\cdot(\frac{1}{3})^{x}=3\cdot\frac{243}{3}\cdot(\frac{1}{3})^{x - 1}=3\cdot81\cdot(\frac{1}{3})^{x - 1}=3\cdot(243)^{\frac{1}{3}}\cdot(\frac{1}{3})^{x - 1}\), but the standard form from the general exponential decay formula \(y=a\cdot b^{x}\) with \(a = 243\) and \(b=\frac{1}{3}\) is \(r(x)=243\cdot(\frac{1}{3})^{x}\).

Answer:

\(r(x)=243\cdot(\frac{1}{3})^{x}\)