Question 6 of 10 , Step 1 of 3
$7 / 15$
Correct
The half-life of gold-194 is approximately 1.6 days.
Step 1 of 3 : Determine $a$ so that $A(t)=A_{0} a^{t}$ 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
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Answer
\(\boxed{a \approx 0.755929}\)
Step 1 ：Given the half-life of gold-194 is 1.6 days, we can set up the equation \( A(1.6) = A_{0} \cdot a^{1.6} = \frac{1}{2} A_{0} \)
Step 2 ：Divide both sides by \( A_{0} \) to get \( a^{1.6} = \frac{1}{2} \)
Step 3 ：Take the 1.6th root of both sides to solve for \( a \), resulting in \( a = \left(\frac{1}{2}\right)^{\frac{1}{1.6}} \)
Step 4 ：Using a calculator, we find \( a = 0.5^{\frac{1}{1.6}} \approx 0.5^{0.625} \approx 0.755929 \)
Step 5 ：\(\boxed{a \approx 0.755929}\)
