Question

Plutonium–238 has a yearly decay constant of 7.9 × 10-3. If an original sample has a mass of 15 grams, how long will it take to decay to 12 grams?

Answer 1

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Answer 2

Heyo! Well your question is posted in the wrong section this is biology so it would be posted in the biology or chemistry section but...here is a link to help you ^^

http://openstudy.com/updates/518d0981e4b0cf6dd8f3e29f

Answer 3

whoops meant this is mathematics section ^^

Answer 4

\[15e^{-.0079t}=12\] solve for \(t\)

Answer 5

Interestingly I forgot about how to do this stuff. (Not the original poster, just trying to refresh my memory.)

An attempt:
\[15e^{-.0079t}=12\]
\[\dfrac{15e^{-.0079t}}{15} = \dfrac{12}{15}\]
\[e^{-.0079t} = \dfrac{12}{15}\]

Which do I just take \(ln()\) of both sides? If I do, why does that work? (e.g. is there a name to the property?)

Answer 6

@opcode yes, you are on the right track. Take the natural logarithm of both sides.

\[\ln(e^x) = x\]

Answer 7

remember, the logarithm to the base \(b\) of \(u\) is some number \(a\) such that \(b^a = u\)

so the log to the base of the base of an exponential just gives you the exponent.

Might be easier to see with common logs.

\[\log_{10} 100 = 2\]because \(10^2 = 100\)

but \(100=10^2\) so \(\log_{10} (10^2) = 2\)

Answer 8

\[ln(e^{-.0079t}) = ln(\dfrac{12}{15})\]
\[\Rightarrow .0079t = ln(\dfrac{12}{15})\]

So I applied that, not sure what a natural logarithm does to a fraction, that has no exponent which in turn has no base \(e\). (Extremely rusty, I know I could just plug it into a calculator, but I am more curious on how it works.)

Answer 9

well, a natural logarithm does the same thing with a fraction it does with any other allowable number. but there is a property of logarithms that can be used if helpful, which is that:

\[\log(\frac{a}{b}) = \log(a) - \log(b)\]

and for completeness:
\[\log(a*b) = \log(a)+\log(b)\]
(the latter being the punch line to a joke about snakes)

Answer 10

because of that property of logs of fractions, you also have this one:

\[\log(\frac{a}{b}) = \log(a) - \log(b) = -(\log(b) -\log(a)) = -\log(\frac{b}{a})\]