Question

Find the half-life of an element, which decays at a rate of 3.411% per day?

Answer 1

\[1-.03411=0.96589\]
solve
\[(0.96589)^x=.5\] via the change of base formula
\[x=\frac{\ln(.5)}{\ln(0.96589)}\] and a calculator

Answer 2

Can you explain how you know to do that?

Answer 3

http://www.wolframalpha.com/input/?i=ln%28.5%29%2Fln%280.96589%29

Answer 4

oh how i know it?
\[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\] it is called the "change of base formula" allows to you solve when the variable is in the exponent

Answer 5

yeah i get that part. i just don't know why you took the decay rate and subtracted it from 1 and all that.

Answer 6

and to decrease a number by \(3.411\%\) you multiply it by
\[100\%-3.411\%=96.589\%=.96589\]

Answer 7

and where did this formula come from: (0.96589)^x=.5?

Answer 8

that is unless you were taught to use
\[e^{-.03411t}\] but that is not really what it says

Answer 9

half life means you end up with half of what you started with
that is why i set it equal to \(\frac{1}{2}\) or \(.5\)

Answer 10

all we were given for decay/growth was:\[q=q _{0}e ^{r*t}\]

Answer 11

ok then go with
\[\large e^{-.03411t}=.5\] so
\[-.03411t=\ln(.5)\] or
\[t=\frac{\ln(.5)}{-.03411}\]

Answer 12

half life, set it equal to one half
then solve for \(t\) as above

Answer 13

that makes more sense. I tried solving for t but got a negative number which didn't look right. but i didn't set .03411 to be negative