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Interpreting change in exponential models: with manipulation

Sal analyzes the rate of change of various exponential models, where the function that models the situation needs some manipulation.

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  • blobby green style avatar for user nakiya booze
    i watched this so many times and still dont get none of it
    (77 votes)
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    • hopper cool style avatar for user Feliz
      you see 1.35^t/6+5=1.35^5times 1.35^t/6, this is logical, because 1.35^5 times 1.35^t/6=1.35^t/6+5
      and (1.35^1/6)^t is also logical, bacause he moved the t outside the parentheses, and so the answer is the same, for example if t=5, 5/6= 0.83, and 1/6 times 5/1=5/6,so this is also logical,
      but I don't really understand the last step
      (2 votes)
  • starky sapling style avatar for user Amalzakelsherif
    At , I repeated the last part many times yet I still don't get it, I just don't understand how 5% is the answer
    (12 votes)
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    • starky ultimate style avatar for user Felicia L.
      The common ratio is 1.051, which means that every time t is increased by 1, M(t) is multiplied by 1.051. Since 1.051 is larger than 0, M(t) will be increasing. 1.051 = 105.1%, and since anything multiplied by 100% is itself, M(t) will increase by 5.1% (105.1-100) every day. Hope this helps!
      (42 votes)
  • starky ultimate style avatar for user Samyak Jain
    At , Did Sal forget to write 1.35^5? Or was that on purpose?
    (16 votes)
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    • blobby green style avatar for user Mic Rofoon
      He did, but it doesn't matter. 1.35^5 = ~4.484. Whether you are raising a sunfish weighing 4.484 grams by 5.1% of its weight or a sunfish weighing 1.35 grams by 5.1% of its weight, you're still increasing its weight by 5.1%. Regardless of whether he played that exponent, the answer to the problem is still 5.1%.
      (3 votes)
  • hopper jumping style avatar for user trinityjqian
    can someone please explain this video in simpler terms? like a step-by-step process? thank you.
    (6 votes)
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    • male robot hal style avatar for user Rushat
      So basically what we want to find in the video is the common ratio for 1 day but we are given the common ratio for t/5 +6 days. By using the properties of exponents. we already know, Sal simplifies it in a way such that the power of the common ratio is 1. Hope you find this helpful.
      (18 votes)
  • blobby green style avatar for user Maple
    at this point the hardest part isnt the math but just reading the problem and answering in the correct units lol
    (14 votes)
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  • piceratops ultimate style avatar for user Garret Cervantez
    Does this mean that the initial mass of the sunfish is 1.35^5 = 4.48...?
    (7 votes)
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  • blobby green style avatar for user Avantika Palayekar
    Anyone who doesn't understand this lesson needs to go back to Unit 6 and fully master Lesson 4 of Unit 6 (may seem frustrating at first, but you will get the hang of it).
    (8 votes)
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  • blobby green style avatar for user lewf
    I feel like this section needs more than just this lesson...
    The techniques required to score on the follow up exercise are not trivial.
    Anyone else?
    (7 votes)
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  • duskpin sapling style avatar for user JJ
    I wanna share this to people who struggle in the exercise for this lesson because I think their explanation for the question is not exactly clear(for me at least).

    For example: A question ask you what t should be if we want a rate of change of (1/2), so you try to do exponent manipulation to turn 1000 ⋅ (1/64)^t into 1000 ⋅ (1/2)^6t.

    1000 ⋅ (1/64)^t
    1000 ⋅ ((1/2)^6)t
    =1000 ⋅ (1/2)^6t



    And you instinctively know or at have some idea what t is, you know the answer but not sure what the process is. And you look for the explanation given by the question which is kinda unclear.




    The method I came up with which i wanna share is: think of original t as t① and think of the t the question trying to ask as t②

    1000 ⋅ (1/64)^t①

    1000 ⋅ (1/2)^6t②

    You know from this, for one power(t①=1) you get a rate of 1/64. And you also know from the exponent manipulation you do earlier, one t② you get a rate of 1/2 which is what we are looking for.

    What should we do to make 1000 ⋅ (1/2)^6t② become true while looking for t②? Algebra, you can express it in an algebraic expression.

    Think of it this way, what 6t② should be in order to make the equation true? One. So if 6t②=1, what is t②? (divide both side by 6) t②=1/6. So there you have it, for power of 1/6 you a get rate of 1/2)

    Whenever you get to this step, and wonder why if its the case, you can try to think it algebraically.





    (Im not a teacher or anything, if i made any mistakes here, feel free to correct me.)
    (7 votes)
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  • starky sapling style avatar for user 𝘽𝘼𝙏𝙈𝘼𝙉
    Can someone give me a few examples of how to solve problems like this? Im kinda-very stuck on it...

    Thankyou!🙂
    (2 votes)
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    • leaf green style avatar for user Katriana
      Here is an example taken from the exercise following this video:
      A sample of an unknown chemical element naturally loses its mass over time.
      The relationship between the elapsed time t, in days, since the mass of the sample was initially measured, and its mass, M(t), in grams, is modeled by the following function:
      M(t)=900⋅(8/27)^t
      The sample loses 1/3 of its mass every ____ days.
      Let's look at the different parts of the equation.
      900 grams is the sample's initial mass.
      Multiplying that by (8/27)^t means that by the end of each day only 8/27 of the amount that was there at the beginning of that day is left. In other words, the sample loses 19/27 of its mass over the course of one day. This is more than 1/3, so it must be that the sample loses 1/3 of its mass in less than one day.
      For the purpose of this exercise, we can mostly ignore the original mass, 900.
      If we can change the common ratio from 8/27 to 2/3 somehow, then we will be able to find out how long it takes for the sample to lose 1/3 of its mass.
      8/27=2/3^3, so:
      (8/27)^t=
      ((2/3)^3)^t=
      (2/3)^3t
      This means that after about 1/3 of a day 2/3 of the mass that was there at the beginning of the day remains. In other words, the sample loses 1/3 of its mass approximately every 0.33 days.
      Hopefully that was helpful. If there is anything unclear or anything you don't understand about this, then you can ask about it.
      (9 votes)

Video transcript

- [Voiceover] Ocean sunfishes are well-known for rapidly gaining a lot of weight on a diet based on jellyfish. The relationship between the elapsed time, t, in days, since an ocean sunfish is born, and its mass, M of t, in milligrams, is modeled by the following function. All right. Complete the following sentence about the daily percent change in the mass of the sunfish. Every day, there is a blank percent addition or removal from the mass of the sunfish. So one thing that we know almost from the get-go, we know that the sunfish gains weight, and we also see that as t grows, as t grows, the exponent here is going to grow. And if you grow an exponent on something that is larger than one, M of t is going to grow. So I already know it's going to be about addition to the mass of the sunfish. But let's think about how much is added every day. Well, let's think about it. Well let's see if we can rewrite this. I'm going to just focus on the right-hand side of this expression. So 1.35 to the t/6 plus five. That's the same thing as 1.35 to the fifth power, times 1.35 to the t/6 power. And that's going to be equal to 1.35 to the fifth power, times 1.35, and I can separate this t/6 as 1/6 times t. So 1.35 to the 1/6 power, and then that being raised to the t-th power. So let's think about it. Every day, as t increases by one, now we can say that we're gonna take the previous day's mass, and multiply it by this common ratio. The common ratio here isn't the way I've written it. Isn't 1.35. It's 1.35 to the 1/6 power. Let me draw a little table here to make that really, really clear. And all of that algebraic manipulation I just did is just so I could simplify this, so I have some common ratio to the t-th power. So t and M of t. So based on how I've just written it, when t is zero, well if t is zero, this is one, so then we just have our initial mass, it's going to be 1.35 to the fifth power. And then when t is equal to one, it's going to be our initial mass, 1.35 to the fifth power times our common ratio, times 1.35 to the 1/6 power. When t equals two, we're just gonna multiply what we had at t equals one, and we're just gonna multiply that times 1.35 to the 1/6 again. And so, every day, we are growing by our common ratio, 1.35 to the 1/6 power. Actually, let me get a calculator out. We're allowed to use calculators in this exercise. So 1.35 to the, open parentheses, one divided by six, close parentheses, power, is equal to, I'll say 1.051, approximately. So this is approximately 1.051. So we can say this is approximately 1.35 times 1.051 to the t-th power. So every day, we are growing by a factor of 1.051. Well growing by a factor of 1.051 means that you are adding a little bit more than 5%. You're adding 0.51 every day of your mass, so you're adding 5.1%. And if you're rounding to the nearest percent, we would say there's a 5% addition to the mass of the sunfish every day.