SOLUTION: Q15. The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nuclei originally present have disintegrated. About

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Question 1007142: Q15. The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nuclei originally present have disintegrated. About how many days can you use the sample?
a. 302
b. 287
c. 312
d. 297

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
WITHOUT MEMORIZED FORMULAS:
t days is t%2F130 half-lives,
After that time, the amount of radioactive nuclei would have been halved t%2F130 times.
So, if you started with an amount A%5B0%5D ,
after that time the amount you would have (halved t%2F130 times) is
A=A%5B0%5D%2A%281%2F2%29%5E%22t+%2F+130%22 .
You could write that as
A%2FA%5B0%5D=%281%2F2%29%5E%22t+%2F+130%22<-->A%2FA%5B0%5D=1%2F2%5E%22t+%2F+130%22<-->A%5B0%5D%2FA=2%5E%22t+%2F+130%22 .
That is what you get just by thinking it through.
It is not a formula that needs to be memorized
(unless you prefer to let other people do the thinking).
So, you could set
A%5B0%5D=%22100%25%22
A=%22100%25%22-%2280%25%22=%2220%25%22
and substituting into A%5B0%5D%2FA=2%5E%22t+%2F+130%22 you would have
%22100%25%22%2F%2220%25%22=2%5E%22t+%2F+130%22
5=2%5E%22t+%2F+130%22
If you take logarithms on both sides of the equal sign, you get an equally valid equation.
You could use logarithms on any base,
but calculators give you logarithms on bases 10 and e .
So, using base 10%29%29%29+%2C%0D%0A%7B%7B%7Blog%285%29=log%28%282%5E%22t+%2F+130%22%29%29
log%285%29=%28t%2F130%29%2Alog%28%282%29%29
log%285%29%2Flog%282%29=t%2F130
t=130%2A%28log%285%29%2Flog%282%29%29
t=highlight%28302%29 (rounded to nearest whole number)

THE FORMULAS:
The irrational number e is a popular base for calculus reasons.
So, from A%2FA%5B0%5D=1%2F2%5E%22t+%2F+130%22 , using logarithms on base e you get
ln%28A%2FA%5B0%5D%29=ln%28%281%29%29-ln%282%5E%22t+%2F+130%22%29
ln%28A%2FA%5B0%5D%29=0-%28t%2F130%29%2Aln%282%29
From there, you get
A%2FA%5B0%5D=e%5E%22-ln%282%29t+%2F130%22<-->A=A%5B0%5D%2Ae%5E%22-ln%282%29t+%2F130%22 ,or using the approximate value ln%282%29=0.693 ,
A%2FA%5B0%5D=e%5E%22-0.693+t+%2F+130%22<-->A=A%5B0%5D%2Ae%5E%22-0.693+t+%2F130%22
Unfortunately, when asked to "show your work",
your teacher may expect to see one of those formulas written out,
just to prove that you are good at memorizing stuff that does not rhyme.