WEBVTT
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Let's look at exponential growth and decay.
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What does it mean when we say something grows exponentially?
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Here is an example. Suppose the population of a certain town
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doubles every year and in the first year, let's call it year 0,
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the population is 100 people. So we would have this:
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when t equals 0, the population is 100.
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In year 1, the population will have doubled since it's doubling every year.
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So when t equals 1, the population is 200 people.
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In year 2, the population will double from the previous year.
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So when t equals 2, the population is 400 people.
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We'll continue this for the first 10 years, so that you can see
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the pattern that develops when we have exponential growth.
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When t equals 3, the population is going to double from 400,
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and it will be 800 people. When t is 4, the population is going to double from 800,
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and you will have 1,600 people in the town.
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When t equals 5, the population doubles from 1,600.
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You'll have 3,200 people in the town in the fifth year.
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When t equals 6, the population will be 6,400.
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When t equals 7, the population will be 12,800.
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When t equals 8, the population will be 25,600.
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When t equals 9, the population will be 51,200.
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Finally when t equals 10, the population will be 102,400 people.
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Notice that after 10 years, the population is over 100,000 people.
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This is characteristic of exponential growth.
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Exponential growth appears to start off slow but eventually becomes very rapid.
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Now we'll do a similar example for exponential decay.
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Lead 210 is a radioactive element with a half-life of 22 years.
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This means that given a certain amount of lead 210,
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every 22 years, half of it will decay.
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So let's suppose that we start off with 80 grams of lead 210.
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Then the first 88 years would look like this:
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When t equals 0, the amount of lead 210 is 80 grams.
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When t equals 22 years, the amount of lead 210 would have been cut in half,
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and we would have 40 grams.
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When t equals 44 years, the amount from the 22 year mark would be cut in half,
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and we would have 20 grams.
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When t equals 66 years, the previous amount from the 44 year mark
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would be cut in half, and we would have 10 grams.
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Finally, when t is 88 years, we would have an amount of 5 grams.
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Both exponential growth and exponential decay can be modeled by this equation:
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A of t equals A sub zero times e to the k t,
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where t is the time,
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k is the growth or decay rate, A sub zero is the amount when t equals zero,
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and A of t is the amount at time t.
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If k is positive, this model is exponential growth,
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and if k is negative this model is exponential decay.
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Now let's do some examples. For example number one, we have
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the size P of a certain insect population at time t, where t is in days,
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obeys the function P of t equals 50 times e to the 0.03 t.
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For part a, we would like the determine the number of insects when t equals 0.
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For this calculation, we simply plug in zero for t,
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and get P of zero equals 50 e to the 0.03 times zero which is 50.
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Now this is e to the zero which has a value of one,
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so 50 times 1 which gives us 50.
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At time zero, there are 50 insects in this population.
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Also notice that this is the same number A sub zero in the formula.
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If I need to know the number of insects at time zero,
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all I simply needed to do was pull A sub zero from the formula.
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For part b, what is the growth rate of the insect population?
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This is simply the value of k. In this problem, k is 0.03.
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If I want to write this as a percent, that would be 3%.
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Now we'll look at the next part of this problem.
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For part c, we would like to determine the population of insects after 5 days.
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For this calculation, we will find P of 5,
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which will be 50 times e to the 0.03 times 5.
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For this calculation, I will first multiply 0.03 times 5
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and get P of 5 equals 50 times e to the 0.15
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On the calculator, I will put in e to the 0.15.
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This is second LN on this calculator. Put in 0.15.
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If your calculator opened parenthesis, go ahead and close them at this point.
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For e to the 0.15, I got approximately 1.16.
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Take this number and multiply it by 50.
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We'll get, for the approximate number of insects, 58.09.
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Now, since the insect population would be a whole number,
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we'll say the approximate number of insects is 58.
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Now we'll look at the next part of this problem.
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For part d, we would like the determine
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the number of days it will take for the insects to reach 400.
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To solve this, put 400 in for P of t.
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We'll have 400 equals 50 times e to the 0.03 t
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To solve this exponential equation, divide both sides by 50.
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We'll get 400 over 50 equals e to the 0.03 t.
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Reducing 400 over 50, we'll have 8 equals e to the 0.03 t.
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To get the 0.03 t down from the exponent position,
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we will take the natural log of both sides,
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giving us the natural log of 8 is the natural log of e to the 0.03 t.
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Using the properties of logarithms,
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I can now take the 0.03 t, and write it in the front of natural log e.
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So natural log of 8 is 0.03 t times the natural log of e.
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Since the natural log of e equals 1,
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this simplifies to natural log of 8 equals 0.03 t.
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To find t divide both sides by 0.03,
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and we'll get the natural log of 8 divided by 0.03 equals t.
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I'm going to go ahead and approximate this on the calculator
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and get natural log 8 divided by 0.03.
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So the approximate number of days is 69.31.
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Rounding to the nearest tenth, we get t is approximately equal to 69.3 days.
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Now we will look at the next part of this problem.
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For part e, we would like to determine
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the number of days that it will take the insect population to double.
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Since the initial amount was 50, this means for it to double,
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P of t would be 100. I'll write 100 equals 50 times e to the 0.03 t.
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To solve this equation for t, first divide both sides by 50.
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We'll have 100 over 50 equals e to the 0.03 t.
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Reducing 100 over 50, we'll have 2 equals e to the 0.03 t.
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To solve this for t we will take the natural log of both sides,
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and this will give us the natural log of 2 is the natural log of e to the 0.03 t.
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This will allow us to take the exponent and put it in the front of natural log e.
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So we'll have natural log 2 is 0.03 t times natural log e.
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Since the natural log of e equals 1,
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this simplifies to natural log 2 is 0.03 t.
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Dividing both sides by 0.03, we'll get
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natural log 2 divided by 0.03 equals t.
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We'll approximate this on the calculator,
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and get natural log 2 divided by 0.03.
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t is approximately 23.1 days.
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Rounded to the nearest tenth, t is approximately 23.1 days.
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Now we'll take a look at the next example.
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For example 2, we have the population of a colony of mosquitos
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obeys the law of uninhibited growth.
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If there are 1000 mosquitos initially and 1300 after 1 day,
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then what is the size of the population after 3 days?
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Since this colony follows the model of uninhibited growth,
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we can model it with the function A of t equals A sub zero e to the k t.
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For this problem, we will have to find the growth rate, k.
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For this, substitute 1300 for A of t,
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1000 for A sub zero, the initial amount,
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e to the k, and t is 1.
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When t is 1, there are 1300 mosquitos.
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To solve this for k, we will have to divide both sides of the equation by 1000.
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This gives us 1.3 equals e to the k.
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Since k times 1 is k, I'll just write e to the k.
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Take the natural log of both sides,
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and get the natural log of 1.3 equals the natural log of e to the k.
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Using the properties of logarithms, we can put k in the front of natural log of e.
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k times the natural log of e is the natural log of 1.3.
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Since the natural log of e is 1,
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we will get k is approximately equal to,
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let's find out the approximation for the natural log of 1.3.
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Natural log of 1.3 is 0.262364
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and this is an approximate value.
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0.262364 for the growth rate.
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From here, we'll go ahead and plug in k into the function,
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and get A of t is 1000 e to the 0.262364 times t.
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To find the size of the colony after 3 days, substitute 3 for t,
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so A of 3 is 1000 times e to the 0.262364 times 3.
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We can approximate this with the calculator,
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and go ahead and multiply the exponent first.
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.262364 times 3 will give me
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A of 3 equals 1000 times e to the 0.787092793.
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Now go ahead and use the answer feature from my calculator,
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and do 2nd, LN, 2nd, ANS to get e to that power.
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Multiply the result by 1000.
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A of 3 is approximately equal to 2197 mosquitos.
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Now we'll take a look at the next example. Example 3:
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Strontium 90 is a radioactive material that decays according to the function
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A of t equals A sub zero times e to the negative .0244 t, where t is in years.
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Assume 800 grams initially. Find the half-life.
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The half-life is the time it takes for half of the material to decay.
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So if we are starting with 800 grams, then we need to know how long it will be
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until there are 400 grams. So it will be 400 for A of t.
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To solve this equation for t, divide both sides by 800.
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So we'll have 400 over 800 equals e to the negative 0.0244 t.
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Reducing 400 over 800, we get
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1/2 equals e to the negative 0.0244 t.
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To solve for t, take the natural log of both sides.
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Natural log of 1/2 is the natural log of e to the negative 0.0244 t.
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Using the properties of logarithms,
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I can take the exponent and put it in the front of natural log e,
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and end up with natural log 1/2 equals negative 0.0244 t times the natural log of e.
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Since the natural log of e equals 1,
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this simplifies to natural log of 1/2 is negative 0.0244 t.
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Dividing both sides by negative 0.0244,
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we'll get natural log of 1/2 divided by negative 0.0244 equals t.
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Using the calculator, we'll get the approximate time for t.
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This will be the natural log of 1/2,
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make sure that when you put in natural log of 1/2, you close the parenthesis
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before you type in divided by negative 0.0244.
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Rounded to the nearest tenth,
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we have that t is approximately equal to 28.4 years.
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Suppose we did not know there were 800 grams initially.
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We can still solve this problem.
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We can substitute for A sub zero 1 unit and then for A of t, we would have 1/2 unit.
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Since half-life is the amount of time it takes for the material to decay half way.
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Substituting 1 for A sub zero and 1/2 for A of t gives me the same equation
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that we have at this point in the solution
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The solution from here would be the same for both problems.
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Practice working these problems until you feel comfortable setting up and solving them.