Question 716601: How long will it take a given quantity of carbon-14 to lose 99.9% of its radioactivity? Half -life(5 730 years) thank you!!
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! An equation which can be used for half-life problems is:
where
t = the number of half-life time units (In this problem the time units of the half-life are years so t represents a number of years.)
 = the initial amount (or the amount at t = 0)
A = the amount after t half-life's
In your problem we are trying to figure out how long it will take for an amount of carbon-14 to lose 99.9% of it radioactivity. If it has lost 99.9% what percent is left? Answer: 0.1% (or 0.001 as a decimal). So if we start with carbon-14 then we are interested in when we reach an amount of  . Putting this expression into the general equation for A we get:
Normally we need to know all but one variable in an equation to solve for the unknown variable. But in this equation, if we look at it carefully, we will see that the 's will cancel out if we divide both sides by it:
Now the equation is down to a single unknown. We can now solve for t. For an equation like this we either use logarithms or find a way to express both sides as powers of the same number. But 0.001 (or 1/1000) is not an obvious power of 1/2, 1/2 is not an obvious power of 0.001 and they are both not obvious powers of some third number. So we must use logarithms to solve this.
Logarithms of any base may be used. If we want the simplest exact expression for the solution we would choose base 1/2 logs (to match exponent's base of 1/2). But we want/need a decimal approximation for the solution, then we should choose a base your calculator "knows", base 10 ("log") or base e ("ln").
Since I suspect you want a decimal, I am going to use ln:
Now we use a property of logarithms,  , which allows us to move the exponent of a log's argument out in front of the log as a coefficient. (It is this very property of logs that is the reason we use them on equations like this. They allow us to move the exponent, where the variable is, to a place where we can "get at it" with regular algebra.) Using this property on our equation we get:
Dividing by ln(1/2):
This is an exact expression for the solution to the equation. For a decimal approximation, get out your calculator. (Just be sure to find the logs first and then divide those logs.
P.S. If we had used base 1/2 logs instead we would have gotten:
which is simpler than the answer we got with ln but harder to turn into a decimal.
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