A 41-gram sample of a substance that's used to detect explosives has a k-value of 0.139. Find the half-life in days.
same formula as before N = N_0 ( e^(-kt)) N_0 = 41 N = 41/2 = 20.5 k = 0.139
N = N_0 ( e^(-kt)) 20.5 = 41 * e^(-0.139)t 0.5 = e^(-0.139)t ln(0.5) = -0.139t
so... 20.5=41^0.139t
yeah just don't forget the e
whats the e?
e is a constant that is (approximately) equal to 2.71828
oh ok
but you don't need to worry about it since it gets eliminated
taking the natural log will "undo" the e
cool
so if we have: ln(0.5) = -0.139t t = ln(0.5)/-0.139 = 5.0 days (depending on how they want you to round it)
my calc gives me 4.98667036374
days?
I think so
y does that look so easy... then u try it and its not
hm
no its just hard for me to learn something new with my ADD
ah, I see
wait im tryin to do one bms... after i divide both sides by N then what
you need to divide both sides by N_0 not N
you should get 0.5 on the left side and e(-something) on the right side
then you have: 0.5 = e(-kt) take the natural log of both sides
ok i got it
how do i get rid of the e?
when you take the natural log you can eliminate e since ln(e) = 1
|dw:1501391672874:dw|
so it would be N=-kt?
the ln(0.5) is still there
ln(0.5) = -kt
oh ok
then you just divide both sides by -k to get t
t = ln(0.5)/k, then just plug into your calculator
if the question asks for k instead it would be k = -ln(0.5)/t
what about the negative before the k?
you can just divide both sides by -1, which will "move" the negative sign over
ln(0.5) = -kt divide both sides by -1 gives -ln(0.5) = kt
** I made a typo earlier
so u still divide by -k?
yeah
ln(0.5) = -kt dividing by -k gives us -ln(0.5)/k = t
how do i get rid of the negative sign bc i dont think it would be -...days
ln(0.5) is also negative, so -ln(0.5) is positive
-ln(0.5) = negative * negative = positive
oh duh lol
im gonna close this and do my other question can u just tell me if im doin it right?
sure
thx
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