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### Course: Digital SAT Math > Unit 3

Lesson 7: Linear and exponential growth: foundations# Linear and exponential growth — Basic example

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

## Want to join the conversation?

- at2:39why does it allow us to doodoo...?(66 votes)
- The power of doodoo is a magnificent thing

It allows the human body to subconsciously do the verb of doodoousing(14 votes)

- dodo mates ......(38 votes)
- trick: just input x values in these equations and check for y(22 votes)
- 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)- The fact that we have a power or exponent (in the form of x), means that this equation is exponential, not linear.(7 votes)

- how can we tell if its exponential or linear(14 votes)
- for exponential x is the exponent(4 votes)

- Can someone help define linear and exponential graphs?? I certainly feel like I have no clue about them.....(5 votes)
- can't we solve this without checking all equations? how can we write the equation on ourselves?(6 votes)
- yeah, just check for the slope and the value of Y when X is zero, then you'llhave the answer(2 votes)

- why am I doing math at this very moment(5 votes)
- 2:40
*Be***SURE***to dodo!*

To get 1600 in SAT...(5 votes) - in this case what`s the common factor??(5 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.