Math  /  Algebra

QuestionThe half-life of silver-108m is approximately 130 years.
Step 1 of 3 : Determine aa so that A(t)=A0atA(t)=A_{0} a^{t} describes the amount of silver-108m left after tt years, where A0A_{0} is the amount at time t=0t=0. Round to six decimal places.

Studdy Solution
Solve for a a . Divide both sides of the equation by A0 A_0 (assuming A00 A_0 \neq 0 ):
a130=12 a^{130} = \frac{1}{2}
To solve for a a , take the 130th root of both sides:
a=(12)1130 a = \left( \frac{1}{2} \right)^{\frac{1}{130}}
Calculate the value of a a and round to six decimal places:
a0.994645 a \approx 0.994645
The value of a a is approximately:
0.994645 \boxed{0.994645}

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