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Determine a so that A(t)=A0at describes the amount of gold-194 left after t days, where A0 is the amount at time t=0. Round to six decimal places

User Miguel Prz
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Answer:

look at the explanation

Explanation:

The equation A(t) = A0 * a^t describes exponential decay or growth, where "a" is the decay or growth factor, and "t" is the time in days. In this case, it's describing the amount of gold-194 left after "t" days.

Since gold-194 is undergoing decay, we know that the decay factor "a" must be between 0 and 1. The general decay formula is given by:

A(t) = A0 * e^(-λ * t),

where λ is the decay constant.

Comparing this with the given equation A(t) = A0 * a^t, we can see that a^t must be equal to e^(-λ * t), which means:

a^t = e^(-λ * t).

Taking the natural logarithm of both sides:

ln(a^t) = ln(e^(-λ * t)),

t * ln(a) = -λ * t,

ln(a) = -λ.

This implies that:

a = e^(-λ).

Given that you want the equation A(t) = A0 * at to describe the decay of gold-194, we need to find the value of "a". In the context of exponential decay, "a" should be equal to e raised to the power of the negative decay constant λ.

a = e^(-λ).

However, you haven't provided the decay constant or any specific information about the gold-194 decay process, so I can't calculate the exact value of "a" for you. If you have the decay constant λ, you can calculate "a" using the equation above.

If you can provide more information about the decay constant, I'd be happy to help you calculate "a" rounded to six decimal places.

User Edward Ford
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