# Finding HALF LIFE (KristaKingMath)

So t is time, that one’s obvious. r is going to be our decay constant, and since we’ve already been given it, we’re just going to be able to plug in for r. Sometimes we’ll be given like the initial amount, the final amount, and time, and asked to find the decay constant. It doesn’t matter what values you’re given, as long as you have three of the four. So we’re going to need values for y, y_0, r and t. If we have three of those four values, we can find the fourth. So t is always going to be time, r is always going to be the decay constant, y_0, this value right here, is always the initial amount of the substance, the amount of the substance that you start with. y is the amount of substance that you’re going to have at the end of the decay process. In this case, because we’re interested in half-life, we’re interested in the amount of time it takes for the original substance to decay to half of its amount.

So this y-value here at the end of the decay process is just going to be half of y_0. Because if we start with the amount y_0, that’s like us starting with 100 grams, here. So if we say y_0 is 100 grams, we want to know how long it takes us to get to 50 grams. And 50 grams is just half of the original amount we started with, so we can say that this is always going to be equal to 1/2 times y_0. Therefore, when we’re dealing with half-life problems, we can always replace y here with (1/2)y_0. Then we set that equal to (y_0)e^(-rt). Now we happen to know that the decay constant is 0.000436, so we can plug that in for r, so we’ll say (1/2)y_0 is equal to (y_0)e^{-} (we have this negative sign from our formula here), (y_0)e^{-0.000436t}.

And now what we can do is divide both sides of our equation by y_0. If we divide both sides by y_0, we’re not changing the value of the equation at all, but we will get y_0 to cancel out of the equation, leaving us with just 1/2 on the left-hand side equal to e^(-0.000436t). Now you can see we’ve just got the single variable left here, which is going to be what we need to solve for, that’s going to give us half-life. In order to solve for t, we need to take the natural log of both sides. Because if we take the natural log of both sides, we end up with ln(1/2) on the left, on the right we have ln(e^(-0.000436t)). And the reason we take the natural log of both sides is because we can get ln and e to cancel out. When you take the natural log of an exponential function, those two things cancel with each other, leaving us with just whatever exponent we have left.

So we end up with ln(1/2) is equal to -0.000436t. Now to simplify this left-hand side we need to keep in mind the law of logarithms, which tells us that when we have the natural log of A divided by B, that’s the same thing as the natural log of A minus the natural log of B. So here we have natural log of 1 divided by 2, that’s the same thing as the natural log of 1 minus the natural log of 2. And we can keep the right-hand side as it is. Now we do that because we know that the natural log of 1 is 0, so this term disappears, and we just have this negative natural log of 2 equals -0.000436t. Well because we have a negative on both sides, we can multiply both sides of the equation by -1, and it’ll cancel out that negative sign. So now we can just say natural log of 2 is equal to this decay constant times t. So if we divide both sides by the decay constant, what we’re left with is t=ln(2)/0.000436. And if we plug this into our calculator here, what we get is a value of about 1,590.

So what we can say is that 1,590 years is the half-life of Radium-226. So no matter what amount of Radium-226 I start with, it could be 1 gram, it could be 100 grams, it could be 100 trillion grams, it doesn’t matter how much I have of Radium-226, it’s going to take 1,590 years, approximately, for that amount to decay to half of the original amount. So I can state that as the half-life of Radium-226. But this problem tells us more than just the half-life of Radium-226. Because notice here, that for half-life, time t, what we ended up with was ln(2) divided by our decay constant. So we could really say that time is equal to ln(2) divided by the decay constant, which is r. And this is actually a formula that we can use every time, for half-life. So what we could say is that half-life is always going to be equal to the natural log of 2 divided by r. And the reason that we can say that, is because when we’re dealing with half-life, remember we’re always going to start with our initial amount of the substance, y_0, since it is half-life, the amount of time it takes for the substance to decay to half of the original amount, we’re always going to substitute here for y, (1/2)y_0.

No matter the values here for r and t, we’re always going to be able to cancel the y_0 from both sides, leaving us with 1/2=e^(-rt). We’re going to end up taking the natural log of both sides, we’re going to end up simplifying the natural log on the left using this law of logarithms here, this natural log of 1 is always going to be 0, so we’re always going to be left with t=ln(2) divided by the decay constant, r. So this formula here that we started with, this is a standard formula that we can use for exponential growth or decay. And so if you want to just memorize this formula, and apply it in lots of different situations, you can do that, and you can go through this half-life process every time.

Or, if you memorize this extra formula here, you can save yourself all this time and just remember that half-life is always going to be ln(2) divided by the decay constant. So if you have that decay constant, r, you just say ln(2) divided by 0.000436, and you’d get the half-life of Radium-226, and you could skip this whole process. Otherwise, if you don’t want to memorize this extra formula, you can always go through this same process, with this exponential decay formula, to get the half-life of the element you’ve been asked about.. 