Question

question 12 • 0f24 step 1 of 3 the half - life of gold - 194 is approximately 1.6 days. step 1 of 3: determine a so that $a(t)=a_0a^t$ describes the amount of gold - 194 left after $t$ days, where $a_0$ is the amount at time $t = 0$. round to six decimal places. answer: to enter your answer (open in new window) 2 points keypad keyboard shortcuts

Explanation:

Step1: Recall half - life formula

The half - life formula for exponential decay is given by \(A(t)=A_0a^t\), and at \(t = T\) (half - life), \(A(T)=\frac{1}{2}A_0\). We know that the half - life \(T = 1.6\) days. So we substitute \(t = 1.6\) and \(A(1.6)=\frac{1}{2}A_0\) into the equation \(A(t)=A_0a^t\).

We get \(\frac{1}{2}A_0=A_0a^{1.6}\). Since \(A_0
eq0\) (because \(A_0\) is the initial amount of the substance), we can divide both sides of the equation by \(A_0\). This gives us \(\frac{1}{2}=a^{1.6}\).

Step2: Solve for \(a\)

To solve for \(a\), we take the 1.6 - th root of both sides. Mathematically, we can rewrite the equation \(\frac{1}{2}=a^{1.6}\) as \(a = (\frac{1}{2})^{\frac{1}{1.6}}\).

First, calculate \(\frac{1}{1.6}=\frac{10}{16}=\frac{5}{8} = 0.625\). So \(a = (\frac{1}{2})^{0.625}\).

Using a calculator, we know that \((\frac{1}{2})^{0.625}=2^{- 0.625}\). We can calculate this value:

\(2^{-0.625}=\frac{1}{2^{0.625}}\). We know that \(2^{0.625}=2^{\frac{5}{8}}=\sqrt[8]{2^{5}}=\sqrt[8]{32}\approx1.542210825\). Then \(\frac{1}{1.542210825}\approx0.645497224\).

Answer:

\(a\approx0.645497\)