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Interpreting time in exponential models

Sal finds the time interval over which a quantity changes by a given factor in various exponential models.

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  • primosaur tree style avatar for user knaseem852
    This section is explained really well and I'm happy to say I understand it well. My questions are:

    1) What does the denominator of 't' in any of the modelled functions mentioned in the video actually mean? What is the purpose/reason/effect of 't' having a denominator ?

    2) Throughout the video in any of the examples, 't' has always been made a value such that the power of 't' and it's denominator would be equal to 1, effectively "removing" the power. what effect would it have on the answer/function if the value of 't' was never a multiple of it's denominator?

    This is my first time asking a question on Khan Academy, hope I have phrased it well enough. Will greatly appreciate a response
    (2 votes)
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    • duskpin sapling style avatar for user Vu
      It means that the common ratio were raised to a fraction of t.

      Let say you have something that grows 50% every 3 days. Notice how it says every 3 days, and not every day. This means for every 3 days it increases by 50%. In term of d days, you would write 1.50^d, but this is not correct, because this means that every day it grows by 50%, so to make it correct, the d days has to be divided by 3, hence we write d/3, which makes it 1.50^(d/3)

      Similarly, if say you have something that grows 50% every day. For d days with the equation 1.50^(d/24) this means it is calculated at every hour (24hrs in a day) not day or 1/24 of a day. So when you you get rid of the 24 to make a whole 1, you are turning it into 1 day and not hour anymore.

      Hope this helps clarifying it.
      (35 votes)
  • leaf orange style avatar for user Vlad Taylor
    At , why does Sal say % after 0.125?
    (5 votes)
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  • aqualine ultimate style avatar for user Eric C
    As a general rule, is it better to use decimals or fractions when doing problems like this? My personal preference is fractions, but could someone please point out the merits of using decimals? For example, if you look at , that table gets a bit messy. Is it better to use fractions in general or are there some cases where decimals are good?
    (4 votes)
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    • aqualine ultimate style avatar for user Alec Traaseth
      It's sometimes hard to immediately see the value of a fraction, where decimals show you very clearly the magnitude of the number, so decimals will be more useful in the real world in situations where you need to know what exactly a number is. I find fractions easier to deal with when manipulating expressions, though, and you always know you'll have the exact value when dealing with fractions where decimals sometimes have repeating numbers and so forth.

      Basically, it really depends on what you're trying to do.
      (7 votes)
  • starky sapling style avatar for user mezal84404
    At about minutes, Sal says that you would be multiplying by 1+4/5. Where did he get the 1 from?
    (4 votes)
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  • duskpin sapling style avatar for user Hayden Ludwig
    What makes Math one of the most hardest subjects in school to learn about? What makes it super important?
    (3 votes)
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  • leaf blue style avatar for user TB
    - Where did I go wrong with this problem? I tried to figure it out like this:

    If the sample lost 87.5% from its initial 320 grams, then the amount left would be 320 - (320*0.875) = 40 grams after 87.5% decrease.

    Now all that is left is to find what input of t provides an output of 40 for the function M(t)

    M(t) = 320 * (0.125)^(t/61.4)
    40 = 320 * (0.125)^(t/61.4)
    I then used properties of exponents and properties of logarithms in order to solve for t and got t = 65... Did I set up the equation correctly or did I make a mistake when solving for t?

    edit: Nevermind, I made a mistake when solving for t. Accidentally misused the property of exponents where a^(b-c) = (a^b)/(a^c). Got the right answer finally.
    (3 votes)
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  • duskpin ultimate style avatar for user Aria
    Where can I find more information about this topic? it always feels intuitive and easy after Sal explains it, yet it took me like 15 minutes to figure out the last problem in the video on my own :(
    (3 votes)
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  • primosaur ultimate style avatar for user NEOVISION
    Uhm.. why do we have to divide " t " by some numbers? we didn't do that before...
    (2 votes)
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  • marcimus red style avatar for user Gloria Kim
    what does one do when no equation is given or the equation given doesn't have the principal amount and instead is a different value? (ex: the 1/2 when asking how much the number of bacteria was halved every _ seconds)? would one just multiply or divide the equation by a certain value to get the value one desires ?
    (1 vote)
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    • duskpin ultimate style avatar for user Victor
      Hey Gloria, one formula for half time is this: a(1/2)^(x/h) where h is the half time and x is the time it takes. If you aren't given the principal amount, then you can simply solve for the variable and you'll get your answer. If you have any questions, feel free to ask!
      (2 votes)
  • starky ultimate style avatar for user troy0bush881
    At , why did he have to put an exponent there? It was already there anyways, so did he just have to identify it to show what he was doing?
    (1 vote)
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Video transcript

- 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.