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Current time:0:00Total duration:6:43

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- After a special medicine is introduced into a petri dish full of bacteria, the number of bacteria remaining in the dish decreases rapidly. The relationship between the elapsed time t, in seconds, and the number of bacteria, N of (t), in the petri dish is modeled by the following function, all right. Complete the following sentence about the half-life of the bacteria culture. The number of bacteria is halved every blank seconds. So t is being given to us in seconds, so let's think about this a little bit. I'll draw a little table here. I'll draw a little table here, so if we have, this is t and this is N of (t) and I'll start with a straightforward t at time equals zero right when we start, this whole thing, if this t is zero, then 1/2 to the zero over 5.5 power, this is 1/2 to the zero, that's all gonna be one and you're just gonna be left with 1000 bacteria in the petri dish. Now at what point do we get to multiply by 1/2? At what point do we get to say 1000 times 1/2? Well in order to say 1000 times 1/2, the exponent here has to be one, so at what time is the exponent here going to be one? Well the exponent here's going to be one, this whole exponent's going to be one when t is equal to 5.5 seconds. So t is 5.5 seconds, and likewise we wait another 5.5 seconds, so if we go to 11 seconds, then this is going to be 1000 times 11 divided by 5.5 is two, so times 1/2 to the second power, so times 1/2, times 1/2. So every five-and-a-half seconds we will essentially have half of the bacteria that we had five-and-a-half seconds ago. So the number of bacteria's halved every 5.5 seconds. And you see it in the formula or in the function definition right over there but it's nice to reason it through and really, really digest why it makes sense. Let's do a few more of these. The chemical element einsteinium-253 naturally loses its mass over time. A sample of einsteinium-253 had an initial mass of 320 grams when we measured it. The relationship between the elapsed time t, in days, and the mass, M of (t) in grams, left in the sample is modeled by the following function. Complete the following sentence about the rate of change in the mass sample. The sample loses 87.5% of its mass every blank days. So this one, instead of saying how much we grew or shrunk by, we're saying a percent change. So if you lose 87.5%, that means that you are left with, left with 12.5%. Which is the same thing, is the same way of saying that you have 0.125% of your mass. So another way of thinking about it is the sample is 0.125 of its mass, of its original mass or how long does it take for the sample to be 0.125 of its mass? And there we could do a similar idea, you see the 0.125 right over here and so I could draw a table if you like, although I think you might guess where this is going. But let me draw a little table here. So, t and M of (t), when t is a zero, M of (t) is 320 and so at what time is M of t going to be 320 times 0.125? Because this, going from this to this, that is losing 87.5% of your mass. Losing 87-- so let me try this way. So this is minus 87.5% of the mass, you've lost 0.875 to get to 0.125. So this, you could just use 0.125 to the first power, so what t do you have to make this exponent equal one? Well t has to be 61.4. 61.4 and where t is in days, 61.4 days. Now you might be tempted to always just pattern match, so, oh what ever's in the denominator, but I really encourage you to think about, cause that's the whole point of these problems. If you just are pattern matching these, well I don't know how helpful that's going to be for you. Let's do one more of these. Howard started studying how the number of branches on his tree grows over time. The relationship between the elapsed time t, in years, since Howard studying the tree, and the number of its branches, N of (t), is modeled by the following function. Complete the following sentence about the rate of change in the number of branches. Howard's tree gains 4/5 more branches every blank years. So gaining 4/5 is equivalent to multiplying, multiplying by, now remember you're gaining 4/5 of what you already are, you're not just gaining the number 4/5, you're getting 4/5 of what you already are. So that's equivalent of multiplying by one plus 4/5, or 9/5. So gaining 4/5 is the same thing as multiplying by 9/5. If I'm 5 years old and if I gained 4/5 of my age, I've gained 4 years to get to be 9 years old, which means I multiplied my age by 9/5. So Howard's tree, you could say, grows by a factor of 9/5 every how many years? Well you can see over here the common ratio is 9/5 and so you're going to grow by 9/5 every time t is a multiple of 7.3. Or I guess you could say every time t increases by 7.3 then this exponent is going to increase by a whole and so you could do that as multiplying again by 9/5. So Howard's tree gains 4/5 more branches every 7.3 years.