Hi and welcome. My name is Anthony Varela. And today we're going to talk about exponential decay. So we're going to be using a formula to represent exponential decay. And we'll use this to solve two problems. Our first example will be about medicine in the bloodstream. And our second example will be about half life of a substance.
So our exponential equation is y equals a times b to the power of x. So we have a base number, b, that is being raised to an exponent power, x. And all of this is being multiplied by a. Now let's compare this to our formula for exponential decay.
So we have y equals a times 1 minus b being raised to the power of x. So in our formula for exponential decay, we have an initial value, a. We have a decay factor, which is 1 minus p. So b is the rate of change. We express this as a decimal.
And it is subtracted from 1 to represent decay. And then we'll see, in our examples, that x is a unit of time. So this is our exponential decay formula that we're going to be using in our following examples. So our first example is 650 milligrams of medicine enters a patient's bloodstream. The medicine dissolves at a rate of 30% per hour.
How much of the medicine is left in the bloodstream after six hours? So we're pulling out our formula for exponential decay. And we're starting with 650 milligrams of medicine. So we know, so far, y equals 650. Now we need to multiply that by our decay factor. So that would be 1 minus 30% expressed as a decimal. So 1 minus 0.30.
And we need to raise this to an exponent power. It would be 6, because we want to know how much medicine is left after six hours. So I'm going to subtract 0.3 from 1 to simplify our equation. So now we need to solve for y equals 650 times 0.70 raised to the 6th power. Well, following our order of operations, we're going to apply the exponent first.
So our exponent of 6 is attached to 0.7. It's not attached to 650. So we're going to apply the exponent first. So we have y equals 650 times 0.117649.
Next we're going to multiply these two numbers then. So y equals 76.47. Now, this means, then, that there is about 76.47 milligrams of medicine in the patient's bloodstream six hours after taking 650 milligrams with that decay factor. So to solve for y, we apply that exponent first, and then multiply by that a value.
So our second example has to do with the half life of a substance. So before I get to our scenario, half life is a length of time. And it's the amount of time it takes for a substance to be cut in half. So, for example, if we have 100 milligrams of a substance, after one half life, we'll have 50 milligrams-- one half of that.
After another half life, we'll have 25-- so on and so forth. So that's what we mean by the half life of a substance. So in our example, a chemical engineer is working with a substance with a half life of 13 years. She has 1,800 milligrams of the substance in the lab. How many years will it take for the substance to decay to 100 milligrams?
So using our formula for exponential decay, we want to know the length of time it will take for this substance to decay to 100 milligrams. So that's our value for y.
What's the expression, then, on the other side of the equation? Well, it's our initial value of 1,800 milligrams. And now, thinking about half life, we said that half life is the amount of time it takes for the amount of a substance to be cut in half. So our decay factor is 1 minus 0.5.
Now our a variable here is x. But x does not equal years. It equals the number of 13 year intervals-- or half lives. So we'll have to account for that once we solve for x. So I'm just going to simplify this equation by subtracting 0.5 from 1.
So this is the equation we need to solve for. What is the value of x? So to solve for x, the first thing that I'm going to do is divide by the 1800. That's because our exponent is attached to 0.5. It's not attached to 1800.
So let's divide that through. So we have 0.0555 repeating. So I'm approximating that to 0.556. This equals 0.5 raised to the power of x. Now to take care of that variable exponent, what I'm going to do is take the log of both sides. So in this example, I'm going to take the natural log.
You can take any log of the same base on both sides because this is an equation. So we have the natural log of 0.5556 on one side of the equation, and the natural log of 0.5 raised to the power of x on the other side of the equation. So I'm going to approximate the natural log on the left side of the equation as negative 2.89.
And now, because we have an exponent inside of a log function, we can bring that exponent out, and then just multiply x by the natural log of 0.5, which is approximately -0.693. Now it's pretty easy to isolate x.
We'll just divide both sides of the equation by -0.693. And we get an x value of 4.17 approximately. Now remember, this is the number of half lives-- not years. Now because one half life is 13 years for this substance, we multiply this number by 13. And a rounding to the nearest year, it will take about 54 years for 1,800 milligrams to decay to 100 milligrams.
So to solve for x, in general, we divide by the a value and then apply the logarithm to both sides. And in doing so, you're going to use a property of logarithms that lets you take an exponent inside of a log function and bring it out to be multiplied by the log. So let's review our lesson an exponential decay.
We have our exponential decay formula, where 1 minus b is a decay factor. And to solve for y, first what you do is you apply the exponent. And then you multiply by a. If you need to solve for x, what you would do first is divide by a and then apply a logarithm to both sides, and use this property then to isolate x. So thanks for watching this tutorial on exponential decay. Hope to see you next time.