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Linear and exponential growth — Basic example

Watch Sal work through a basic Linear and exponential growth problem.

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  • aqualine sapling style avatar for user Katie
    One of the practices of the topics gives a question in like this:

    P = A(1-n/100)^x

    Then, it asks you if this is linear or exponential, and why, and gives you several multiple choice options, the correct one being "This is an exponential equation because x is an exponent of the constant."

    My question is this: why is it an exponent of a constant if the only exponent ("x") modifies something that varies based on an input (1- n/100)? How is that a constant? Or is it because "x" modifies "A"?
    (15 votes)
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  • blobby green style avatar for user Taya Wongsalong
    how can we tell if its exponential or linear
    (13 votes)
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  • duskpin seed style avatar for user Fluff-ball
    Can someone help define linear and exponential graphs?? I certainly feel like I have no clue about them.....
    (6 votes)
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  • duskpin ultimate style avatar for user Yisae
    at why does it allow us to doodoo...?
    (1 vote)
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  • blobby green style avatar for user Andy Steckelberg
    Sal 350(2)^3 isn't 2100 its 2800
    (1 vote)
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  • blobby green style avatar for user Cassandra R. Grady
    Is it easy to percentage or fractions?
    (1 vote)
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  • male robot johnny style avatar for user JahGlizz
    One of the practices of the topics gives a question in like this:

    P = A(1-n/100)^x

    Then, it asks you if this is linear or exponential, and why, and gives you several multiple choice options, the correct one being "This is an exponential equation because x is an exponent of the constant."

    My question is this: why is it an exponent of a constant if the only exponent ("x") modifies something that varies based on an input (1- n/100)? How is that a constant? Or is it because "x" modifies "A"?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Shaikh Jawad Bin Islam
    Isn't it possible to just "eye ball" the answer in this case? I understand that this is risky in some cases but saving time is also a necessity in this test.
    (0 votes)
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    • blobby green style avatar for user Steven Spehar
      Saving time may be valuable towards this test, but you also want to make sure that you have the answer. Everyone hates the feeling of eyeballing or guessing on it and being just a little bit off. It feels disappointing to know you could’ve gotten full credit for the question if you solved for it.
      (2 votes)

Video transcript

- [Instructor] We're asked which of the following equations relates y to x for the values in the table above? Pause this video and see if you can have a go at that. All right, so we could just go point by point or coordinate by coordinate, depending how you wanna think about this. So, first of all, when x is equal to one, y is equal to zero. So is that true right over here? When x is equal to one, this is gonna be 1/2 plus one, which is not equal to zero. That would be three halves, so we can rule this one out. When x is equal to one, let's see, two times one is two, minus two is equal to zero. So this one is looking good so far, just based on that first x-y pair. And let's see when x is equal to one here, 1/2 to the first power is 1/2, times two is one, not zero. So we can rule this one out. All right, when x is equal to one here, this is two to the first power minus two, that's two minus two, that is equal to zero. So this is looking good as well. We've already been able to remove two choices. Now, let's see, when we have two and two, let's just go to this one right over here. Two times two is four, minus two is equal to two. So this one's looking good on that second x-y pair. Now, let's see, two to the second power is four, minus two, that is still, that is also equal to two. So this one's looking good. Now, let's think about three, six. So two times three is equal to six minus two, which would be equal to four, not six. So we could rule this one out. And hopefully this will work for this one over here. So let's see, two to the third power is equal to eight minus two is indeed equal to six. So we like this choice right over here. Another way that you might have approached it is you could have said, "Look, this is not a linear relationship." Because every time we're increasing x by one, here we're increasing by two, here we're increasing by four, here we are increasing by eight, here we're increasing by 16. If it was a linear relationship, every time we increased x by one, we would increase y by the same amount every time. And so we know it's not linear, so you could rule out A and B. And you can also see that as x is increasing, y is increasing. So you know it would not be this exponential right over here, because here, as x increases, 1/2 to larger and larger powers would become a smaller and smaller value, y would decrease. So that would also allow you to deduce that D is the choice.