Question 6 of 10 , Step 1 of 3
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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.

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$a \approx 0.755929$

Steps

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 $a = {(\frac{1}{2})}^{\frac{1}{1.6}}$

Step 4 ：Using a calculator, we find $a = {0.5}^{\frac{1}{1.6}} \approx {0.5}^{0.625} \approx 0.755929$

Step 5 ： $a \approx 0.755929$