Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (lgbasallote):

APPLICATION OF FIRST ORDER D.E. (EXPONENTIAL DECAY) scientific studies show that DNA deteriorates over time. if excavations and diggings of fossils indicate that only 1.2% of DNA survive after 25 million years, what percent of an original sample would have been lost after 10 million years? Answer: 82.95% i was able to arrive with the answer but i had to "cheat" to get it. can someone please show me how to work this problem? i may have done something wrong

OpenStudy (lgbasallote):

@Valpey ?

OpenStudy (unklerhaukus):

first order decay

OpenStudy (lgbasallote):

okay here's what i did \[\ln (\frac{x_o}{0.012x_o}) = 25k\] is that right?

OpenStudy (unklerhaukus):

i wouldn't start like that

OpenStudy (lgbasallote):

then what

OpenStudy (unklerhaukus):

\[X(t)=X_0e^{-k t}\]\[\frac{X(t)}{X_0}=e^{-k t}\]\[\ln\frac{X(t)}{X_0}=-kt\]\[\ln\frac{X_0}{X(t)}=kt\]

OpenStudy (unklerhaukus):

\[k=\frac1{t}\ln\frac{X_0}{X_t}\]\[k=\frac1{25\times 10^6}\ln\frac{1}{0.012}\]

OpenStudy (lgbasallote):

xt is?

OpenStudy (unklerhaukus):

\(X_t\) should be \(X(t)\) sorry

OpenStudy (unklerhaukus):

so have for found \(k\)

OpenStudy (lgbasallote):

so it's \[\ln (\frac{x_o}{x(t)})?\]

OpenStudy (unklerhaukus):

what you just wrote is \(=kt\)

OpenStudy (lgbasallote):

should it not?

OpenStudy (lgbasallote):

so in a way what you did is just the same as i did right? \[\large \frac{\ln (\frac{x_o}{0.012x_o})}{25} = k\] correct?

OpenStudy (lgbasallote):

@UnkleRhaukus

OpenStudy (unklerhaukus):

yeah that is right if your working in millions of years

OpenStudy (lgbasallote):

yup...so k = 0.1769 /million years

OpenStudy (lgbasallote):

what do i do next?

OpenStudy (unklerhaukus):

now just use the first equation i wrote ,up the page to find how much DNA will remain after 10million years

OpenStudy (unklerhaukus):

a take this away from 100%

OpenStudy (lgbasallote):

that's actually my question...why do i use \[\ln (\frac{x_o}{x_o - ax_o})\] where a is the percent. when in the other one i just used \[\ln (\frac{x_o}{0.012x_o})\] what's the differece?

OpenStudy (unklerhaukus):

where did you get this from/\[\ln \left(\frac{x_o}{x_o - ax_o}\right)\]

OpenStudy (unklerhaukus):

\[X(t)=X_0e^{-k t}\] \[X(t)/X_0=e^{-0.18t}\] \[X(t)\%=e^{-1.8}\]

OpenStudy (unklerhaukus):

percentage lost: \[1-X(t)\%\]

OpenStudy (lgbasallote):

yes @UnkleRhaukus my question is how do i know when to do 1 - x(t) or just simply x(t)

OpenStudy (unklerhaukus):

x(t) is how much is remaining 1-x(t) is how much has decayed

OpenStudy (lgbasallote):

OHHHH I SEE

OpenStudy (unklerhaukus):

eightee-free

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!