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Mathematics 9 Online
OpenStudy (anonymous):

can anyone help me with exponential decay?

OpenStudy (anonymous):

why not?

OpenStudy (anonymous):

\[Q=Q_0 e^{-rt}\]

OpenStudy (anonymous):

Situation: a 42 gram sample of substance that's used to treat thyroid disorders has a k-value of 0.124 Find the substance's half-life, in days. Round your answer to the nearest tenth.

OpenStudy (anonymous):

i am assuming "k-value" is r, so the equation is \[Q=Q_0 e^{-.124t}\] and half life means how long before you have half of your original amount. it doesn't matter what the original amount is.

OpenStudy (anonymous):

so set \[e^{-.124t}=\frac{1}{2}\] and solve for t.

OpenStudy (anonymous):

\[e^{-.124t}=\frac{1}{2}\] \[-.124t=ln(\frac{1}{2})\] \[t=\frac{ln(\frac{1}{2})}{-.124}\]

OpenStudy (anonymous):

i get 5.59 rounded. to the nearest tenth it would be 5.6 hope the steps are clear. if not let me know.

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

oh wait. i did not use the 42 grams part, so maybe i don't know what k-value is. can you tell me what k-value means exaclty?

OpenStudy (anonymous):

that answer was correct

OpenStudy (anonymous):

whew so k-value is the rate of decay and the 42 was a red herring. good!

OpenStudy (anonymous):

5.6

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