Question

Plutonium-240 decays according to the function

Here's the picture http://media.apexlearning.com/Math/200707/09/1ac96390-8070-493a-aac0-4526746ab16c.gif

where Q represents the quantity remaining after t years and k is the decay constant, 0.00011... How long will it take 24 grams of plutonium-240 to decay to 20 grams?

A. 1.06 years
B.80.11 years
C. .00048
D. 1,600 years

Can someone help me?

Answer 1

@zepdrix can you help me?

Answer 2

\(\large\rm Q(t)\) represents the amount of plutonium remaining at time \(\large\rm t\).
\(\large\rm Q_o\) represents the amount of plutonium that we start with.
And we're given the decay constant \(\large\rm k=0.00011\)

So we're given a starting amount of plutonium, \(\large\rm Q_o=24\)
and an ending amount, \(\large\rm Q(t)=20\)

Let's plug all of this information in,
and see if we can solve for \(\large\rm t\).

Answer 3

\[\large\rm Q(t)=Q_o e^{-kt}\]becomes\[\large\rm 20=24 e^{-0.00011t}\]

Answer 4

Let's isolate the exponential, dividing by 24,\[\large\rm \frac{20}{24}=e^{-0.00011t}\]

Answer 5

What next?
Any ideas?

How do we get the t `out of the exponent`?
Any special tools that will help us? :)
Hmm

Answer 6

is the answer D?

Answer 7

@zepdrix

Answer 8

Yes :o

Answer 9

how did you get the answer? my answer was 1657 so i guessed the answer was D

Answer 10

@zepdrix please explain

Answer 11

Yes 1657.5 is correct :)
I don't know why they rounded so much of that off...
but that's the only option that makes any sense :D