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Mathematics 10 Online
OpenStudy (anonymous):

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OpenStudy (e.mccormick):

So you do not have a specific half life formula, and need to find it.

OpenStudy (e.mccormick):

OK, so show the work.

OpenStudy (e.mccormick):

Sure.

OpenStudy (e.mccormick):

Looks pretty good. So far I got to: \[100=A_0e^{-x30};\\ 30=A_0e^{-x120}\implies\\ 100/e^{-x30}=A_0\implies\\ 30=(100/e^{-x30})e^{-x120}\implies\\ ln(30)=ln((100/e^{-x30})e^{-x120})\implies\\ ln(30)=ln(100)-ln(e^{-x30})+ln(e^{-x120})\implies\\ ln(30)=ln(100)+x30-x120\]

OpenStudy (e.mccormick):

Yah, I just took it to finding x, or your k, and I got the same number.

OpenStudy (e.mccormick):

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.

OpenStudy (e.mccormick):

Not expecially. I need to work on calc more these days.

OpenStudy (e.mccormick):

np. Have fun!

OpenStudy (e.mccormick):

Also got the same \(A_0\)

OpenStudy (e.mccormick):

And with those two, the rest is easy.

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