WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we are going to go over exponential growth and decay.
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This topic was introduced briefly in a previous lecture on exponential equations.
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And now, we are going to look at it in greater depth and work on some sample problems.
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First, decay: there are actually two formulas that we are going to go over for decay, and two for growth.
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The first decay formula talks about a situation where you have a quantity that decreases by a fixed percentage each year.
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The equation for this situation is y = a(1 - r)^t.
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This gives the amount, y, of the quantity after t years; so a is the initial amount that you begin with;
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r is the percent decrease; and in order for this formula to work correctly, you need to express it as a decimal.
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y is the amount that you end up with, the quantity after t time; and t is the time that has passed.
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A typical application for this formula would be in finance.
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For example, if you have $1,000 in your bank account, and you remove 40% of whatever you have in that account each year,
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and you are trying to figure out how much you would have left in 3 years, you could use this formula.
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So, I said you began with $1,000; therefore, a = 1,000 dollars--that is how much money you started out with in your savings account.
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The percent decrease is going to be 40%, because you are taking that amount out each year (or .4).
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And we are trying to figure out how much money you are going to have left after 3 years,
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so the unknown is the amount of money that you have left after 3 years; and the time is 3 years.
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So, trying this problem out using this formula: we are going to take y =...and y is our unknown;
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a is 1,000; times 1 minus...and it is .4, expressed as a decimal, raised to the t power (which here is 3).
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That gives me y = 1000(0.6)³; .6² is .36, times .6 is .216; so this leaves me with y = 1000(.216).
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And multiplying that out, I would find that I would have 216 dollars left in my account after 3 years of removing 40% of the money in the account.
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So again, this decay formula is the general formula often used for applications, talking about financial applications, interest...
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Actually, interest would be growth...but interest payments...decreases; we are talking about decreases.
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The scientific model of decay is slightly different.
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And in a lot of applications in science, the formula that is used is exponential decay: y = ae^-kt.
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We talked previously about how base e is often used in scientific applications; we see it popping up here again.
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And as in the last formula, y is the amount that you end up with after t years, whereas a is the initial amount.
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This time, instead of having a fixed percent that it decreases by, we use a constant.
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And the constant depends on the situation.
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A typical situation would involve radioactive decay; and with radioactive decay, or any type of decay, you will be given a constant.
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There might be a certain isotope or a certain compound that decays according to one k.
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And then, you look at a different compound, and it might decay a lot faster; and then the value of k would be different.
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So, just to review what the graph would look like when we are talking about an exponential decrease:
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the shape of the graph would be roughly like this, over time.
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You see here: as time moves forward, the amount that you have left is getting smaller and smaller and smaller and smaller, and approaching 0 eventually.
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This formula, as I mentioned, is important when working with radioactive substances and thinking about radioactive decay.
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And you may have heard of the concept of half-life.
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Half-life is the amount of time it takes for half of the material to decay away.
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So, let's say I have 10 grams of a radioactive substance, and it takes 3 months for it to decay until there are only 5 grams left.
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The half-life of that substance would be 3 months, the amount of time it takes for half of it to be gone.
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And let's look at an example of this: let's say that I have a radioactive compound that decays exponentially according to this formula.
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And k equals 0.1 for a certain radioactive substance; and it follows exponential decay; and I want to know the half-life.
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All right, let's look at this formula and think about how this will work.
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I have y = ae^-kt, and I am not given the original amount, which seems like it would be a problem;
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but it is actually not, and here is why: think about what half-life is saying.
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It is saying that half of the original amount is left.
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So here, the amount we are left with, y, equals half the original amount, or 1/2a.
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Knowing that, I can set this up as follows: 1/2a = ae^-kt.
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Once I have this set up like this, the a's will cancel; I am going to divide both sides (I am trying to get rid of the a on the right) by a.
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That cancels; this cancels; and I just am left with 1/2 = e^-kt.
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Recall, earlier on, when we were working with exponential equations, we talked about how it is possible to solve these by taking the log of both sides.
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So, let's go ahead and fill in what we have for k, which is -.1.
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And what we are looking for is the time--the amount of time it takes for half of the substance to decay.
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So, we have it down to this point; and what I can do is take a log of both sides.
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And since I am working with base e, I am going to go ahead and work with the natural log,
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although you actually could use any log, as long as it has the same base.
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That is going to give me ln(1/2) = ln(e^-.1t).
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Because these are inverses, the ln(e)...these are essentially going to cancel each other out.
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And on the right, I am going to be left with -.1t.
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On the left, I am going to actually rewrite this as ln(2^-1); and recall that 2^-1 is the same as 1/2.
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All right, let's go on up here to finish this one out.
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ln(2^-1) = -.1t: recall, from the rules of working with logs, that I can rewrite this as -1ln(2), or just -ln(2), equals -.1t.
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I am going to divide both sides by -.1; the negatives are going to cancel out, and this is going to give me ln(2)/.1 = t.
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So, I am rewriting this like this, ln(2) divided by .1.
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You can use your calculator and the natural log button on there to find out what ln(2) is, divided by .1.
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And that actually comes out to approximately 7 months; so it would take 7 months for this compound to decay until there is only half left.
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Therefore, the half-life is 7 months.
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That was decay; now we are talking about growth.
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OK, so again, there are two formulas, and they are very similar to the decay formulas.
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Here, we are talking about growth, so for a quantity that increases by a fixed percentage each year, the equation...
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this actually should be a plus right there...y = a(1 + r)^t gives the amount, y, of the quantity after t years.
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So, a is the initial amount, just like we talked about with decay.
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And r is the percent increase, and we need to express that as a decimal.
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And what we are looking for is the final amount, the amount that we are going to end up with.
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And this is the general formula; so this is the model that you might see used again when working with money or financial-type problems.
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Let's look at an example: let's say that you had $500 in your bank account, so a = 500.
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And you receive 10% interest per year on that.
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I am going to rewrite that in a decimal form: .1.
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And what you want to know is how much money you are going to end up with after 3 years.
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So, rewriting this as y = a(1 + r)^t, this is going to give me y = 500 (that is what I started out with);
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and this is 1 + .1, raised to the t power; and in this case, t is actually 3--let's go ahead and make that 3.
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y = 500(1.1)³
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And you could use your calculator to determine that 1.1 raised to the third power is equal to 1.331.
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Multiplying that by 500, you would end up with 665 dollars and 50 cents in your account after 3 years, if you are receiving 10% interest on that $500.
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So, you can see that this is a really useful formula.
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The scientific model of growth is analogous to the scientific model of decay that we talked about before.
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Notice here that now we have a positive exponent; with the decay formula, y = ae...it was -kt; now we have a positive exponent.
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And the constant k depends on the situation.
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You might be asked to find it, if they give you what y is, and a, and t; or you may be given the k and then asked to figure out something else.
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An important application here in science is in population growth.
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And systems that follow this model of growth increase exponentially, so their graph is going to look approximately like this.
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So, let's consider an example where you have a population of 1,000 people.
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There are 1,000 people living in a town, and this town's population is increasing according to this exponential model.
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And k is 0.2: we want to know the number of people that we are going to end up with in 10 years.
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So, using this formula, y = ae^kt, we don't know y; we know that a is 1,000; e; k is .2; and time is 10 years.
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This gives me y = 1,000e...0.2 times 10--that is just going to give me 2.
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Now, I can figure this out; I could use my calculator, or you might remember that e is 2.7182.
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So, if I square that, I will get approximately (for e²) 7.389.
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Multiply by 1,000: I am going to end up with 7,389 people living in the town 10 years down the road.
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OK, so that was the four equations that we are working with today: two each for growth and decay.
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So, let's try some examples: A computer is purchased for $2,000, and it is expected to depreciate at the rate of 20% per year.
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Depreciate means it is losing value: so you bought your computer, and each year it is worth less,
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which is frequently the case with objects that you buy.
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Let's think about what formula we are going to use.
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Since this is decreasing, we know that we need to use a decay formula.
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And since it is decreasing at a certain percentage per year, I am going to use the general decay formula, which is y = a(1 - r)^t.
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My initial amount is $2,000; that is what I purchased it for.
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And the rate of depreciation is 20%; in decimal form, this is .2.
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I want to know its value after a certain amount of time; and the amount of time is 3 years.
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With this in mind, I can just go ahead and work this out.
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y is what I am looking for; a is 2,000; times 1 - .2, raised to the third power.
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Therefore, y = 2,000...1 minus .2 is .8, to the third power; you can go ahead and work this out on your calculator;
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and you will find that it comes out to .512, times 2,000, is 1,024 dollars.
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So, this computer, in three years, at this rate of depreciation, is going to be worth $1,024.
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Now, we are talking about a stock that is expected to increase in value.
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Since we are talking about increasing, we are talking about growth.
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And this is at the rate of 45% per year.
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Since I am talking about a steady increase that is based on percent per year, I am going to use the general growth formula, not the scientific exponential one.
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So, I am going to use this formula: y = a(1 + r)^t.
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And I want to know how long it will take for the stock to triple in value.
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So, you have to think carefully about how to set this up.
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I know the rate: the rate is given as 45%, which is equal to .45.
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And I am asked how long it will take for the stock to triple in value; so I want to know
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when the value I am going to end up with, y, is 3 times the initial value, 3a.
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So, I have an initial value, a: it doesn't matter what that value is.
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What I want to know is when it is going to be 3 times whatever a is.
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So, if you just keep going on this, you see that the a will drop out.
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So then, I set this up as...let's go ahead and start it over here...for y, I am going to substitute 3a.
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This equals a; I don't know what a is--I just put a.
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1 + .45...and I know that what I am looking for, actually, is t: so, I just leave this as t.
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When I divide, if I want to move this a over here, I am going to divide both sides by a; and the a's will conveniently drop out,
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leaving us with 1 + .45, raised to the t power: so 3 = 1.45^t.
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Recall that, when you are working with exponential equations with different bases
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that you can't easily get to become the same base, you can just take the log of both sides.
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And you can use any base log you want; but two convenient ones are the natural log and the common log.
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So, I am going to go ahead and take the common log, the base 10 log, of both sides.
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log(3) = log(1.45^t).
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Using properties of logs, I can rewrite this as t times log(1.45); and I am looking for t.
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I want to isolate t, so I am going to make this log(3), divided by log(1.45), dividing both sides by log(1.45).
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And remember that these are just numbers; they have a value--I am going to work with them and move them around, just like I would any other numbers.
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I am rewriting this so that the t is on the left--a more common form.
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This is something you can then use your calculator to figure out.
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And if you take the log of 3, divided by log base 10 of 1.45, you will get approximately equal to 3 years.
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So, even though I didn't know what my stock was worth to start with, it didn't matter.
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All I wanted to know is how long it is going to take whatever amount I have to triple; and it is going to take 3 years.
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Example 3: A medicine disintegrates in the body at a steady rate.
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We are talking about disintegration, which is a type of decrease; so we are talking about decay.
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It decays based on the equation y = ae^-kt.
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This is using the scientific model of decay with base e, where k equals 0.125, and t is in hours.
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Find the half-life of this medicine.
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We are using this formula; let's rewrite it here; and we want to find the half-life.
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So, we are looking for the half-life, which is often written this way: t^1/2.
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And I don't know how much medicine we are starting out with, but it doesn't matter.
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It is similar to the last one we worked out, where it doesn't matter what you are starting out with.
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You are just looking for a change.
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I know k; that is given; and I know that what I am going to end up with is half of the initial.
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Since I am looking for half-life, after this amount of time, y is going to be equal to half of the amount that we started out with, 1/2a.
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I am coming over here and setting this up, substituting 1/2a for y, equals ae, and then I have -.125 as k; and what I am looking for is t.
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If I divide both sides by a, the a will drop out.
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This gives me 1/2 = e^-.125t.
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One way to solve this is to take the log of both sides; I am going to go ahead and take the natural log of both sides; this is ln(1/2) = ln(e^-.125t).
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Because these are inverses, I am going to end up with -.125t.
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To go further with this, I am going to rewrite this left part, ln(1/2), as ln of 2 raised to the -1 power,
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just like the example we worked out a few slides ago, equals -.125 times t.
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I can bring this out in front, which would be -1ln(2), or just -ln(2), equals -.125t.
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I am going to divide both sides by -.125, and this equals t; so these negatives...a negative divided by a negative
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is going to become a positive, so I come up here and get ln(2) divided by .125 = t.
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Now, I can find the natural log of 2 and divide that by .125, using a calculator.
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And it turns out that this is approximately equal to 5 hours.
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Therefore, the half-life of this medication, given this constant, is approximately 5 hours, based on using this model for exponential decay.
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Example 4: The population growth of a city can be modeled exponentially with a constant of k = 0.01.
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The current population is 100,000; what will it be in 100 years?
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Since this is exponential growth, and we are given a constant, I am going to use the scientific model for growth, that formula, y = ae^kt.
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All right, I have been given a k; this is 0.01; and I have been given the original population, which is 100,000.
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I have been given a t of 100 years, and I am asked to find y.
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I want to know what the population is going to end up being after 100 years.
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Therefore, this becomes y = 100,000 times e; and that would be a k of .01, times a t of 100.
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Therefore, this is y = 100,000 times e; and this is just going to be .01 times 100, so that is going to give me 1.
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Recall that e is 2.71828, so instead of e to the first power, it is just e, which I know the value of.
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Then, I go ahead and multiply this times 100,000 to get that 271,828 will be the population in 100 years.
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Again, this was simply a problem involving the use of the scientific model for growth.
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That concludes this lesson on Educator.com on exponential growth and decay; thanks for visiting!