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So you do not have a specific half life formula, and need to find it.
OK, so show the work.
Sure.
Looks pretty good. So far I got to: \[100=A_0e^{-x30};\\ 30=A_0e^{-x120}\implies\\ 100/e^{-x30}=A_0\implies\\ 30=(100/e^{-x30})e^{-x120}\implies\\ ln(30)=ln((100/e^{-x30})e^{-x120})\implies\\ ln(30)=ln(100)-ln(e^{-x30})+ln(e^{-x120})\implies\\ ln(30)=ln(100)+x30-x120\]
Yah, I just took it to finding x, or your k, and I got the same number.
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So you took the tacktic that it was solving two equations with two unknowns and an added twist involing e and ln. All 100% the right way to go.
Not expecially. I need to work on calc more these days.
np. Have fun!
Also got the same \(A_0\)
And with those two, the rest is easy.
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