# What is the half life of a 500 gram sample that decays to 62.5 grams in 639,000 years?

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Half-lives are not as tricky as books often make it seem. Many give equations of the form:

`m(t) = m_0e^(-rt)`

where `m` is the mass after a period of time, `m_0` is the initial mass, `t` is time, `r` is a constant, and `e` is the base of the natural logarithm.

Now, I will point out that you, in fact, can use any number instead of `e` to make the equation work. In fact, after I show you how to solve it with the standard equation, I'll show you (in my opinion) a much easier method.

So, to find the half-life with the numbers we have above, we solve in two steps:

1) Find `r` in the equation

2) Find `t` such that `m(t) = m_0/2`

So, let's start by finding `r`. Let's plug the values we have above into our equation:

`62.5 = 500*e^(-r*639000)`

We simplify by dividing by 500:

`0.125 = e^(-r*639000)`

Now, we take the natural log of both sides to bring down the expression in the exponent:

`ln(0.125) = -639000r`

Finally, we divide both sides by -639,000 to find r. Of course, we'll need a calculator for this step:

`3.254*10^-6 = r`

Now, we can move on to the second step and solve for the `t` at which our mass is half of the original mass. Let's substitute the values into the equation again, but now we're going to solve for t when `m(t) = m_0/2 = 500/2 = 250`:

`250 = 500e^(t*-3.254*10^-6)`

Now, we again divide by 500 to isolve the exponential expression:

`1/2 = e^(t*-3.254*10^-6)`

Now, we take the natural logarithm again:

`ln(1/2) = t*(-3.254*10^-6)`

Finally, we divide by -3.254*10^-6 to get our answer:

`213013 = t`

Now, this answer is a bit off due to the rounding we did. However, if you were to never round, you would get the absolute correct answer, but it would take a bunch of space on the page.

Now, when you're dealing with half-lives, it's easier to use the following equation:

`m(t) = m_0 2^(-rt)`

This equation would allow you to perform the same calculations to find r; however, when finding the half life, you can skip the logarithm and exponential part because you only need to solve for when the exponent becomes -1! In other words, you solve this equation:

`-rt = -1`

This is true because when you get the value for time that makes the above equation true, you get the following result:

`m(t) = m_0*2^-1 = m_0/2`

Finally, for this problem, there is a special case. If you noticed way back in our first step of the first method, we got the following result:

`0.125 = e^(-r(639000))`

If you'll notice, 0.125 is one-eighth, so we get the following equation:

`1/8 = e^(-r(639000))`

Notice, now that `1/8` is equivalent to `(1/2)^3` :

`(1/2)^3 = e^(-r(639000))`

The significance here is that, based on the rules for exponents, the power of `e` is triple what it would be to get the half life. We can actually take each side to the 1/3 power to solve for the half life directly:

`1/2 = e^(-r(639000)/3)`

The expression in the exponent, remember is of the form -rt. Therefore, if we simplify the following, we get our correct answer for the half life:

`639000/3 = 213000`

So, our half life is 213,000 years. Pretty much the same as what we got in our above analyses. I hope one of these three methods makes sense to you!

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