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
@Valpey ?
first order decay
okay here's what i did \[\ln (\frac{x_o}{0.012x_o}) = 25k\] is that right?
i wouldn't start like that
then what
\[X(t)=X_0e^{-k t}\]\[\frac{X(t)}{X_0}=e^{-k t}\]\[\ln\frac{X(t)}{X_0}=-kt\]\[\ln\frac{X_0}{X(t)}=kt\]
\[k=\frac1{t}\ln\frac{X_0}{X_t}\]\[k=\frac1{25\times 10^6}\ln\frac{1}{0.012}\]
xt is?
\(X_t\) should be \(X(t)\) sorry
so have for found \(k\)
so it's \[\ln (\frac{x_o}{x(t)})?\]
what you just wrote is \(=kt\)
should it not?
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?
@UnkleRhaukus
yeah that is right if your working in millions of years
yup...so k = 0.1769 /million years
what do i do next?
now just use the first equation i wrote ,up the page to find how much DNA will remain after 10million years
a take this away from 100%
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?
where did you get this from/\[\ln \left(\frac{x_o}{x_o - ax_o}\right)\]
\[X(t)=X_0e^{-k t}\] \[X(t)/X_0=e^{-0.18t}\] \[X(t)\%=e^{-1.8}\]
percentage lost: \[1-X(t)\%\]
yes @UnkleRhaukus my question is how do i know when to do 1 - x(t) or just simply x(t)
x(t) is how much is remaining 1-x(t) is how much has decayed
OHHHH I SEE
eightee-free
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