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Here is how we can write an equation for an exponential function from a table of values: 1. Determine the common ratio. For example, if we see that every time x increases by 1, y is multiplied by 2, then the common ratio is 2. 2. Find the initial value of the function, or the y-intercept. This is the y-value when x=0. 3. Write the equation in the form y = ab^x, where a is the initial value and b is the common ratio.
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- how does he divide 144 by 3 IN HIS HEAD?(11 votes)
- Well, what's 24/3? 24/3 is 8, we can do this in our head. Why is this important? 3*10=30, we know that 30+30+30+30=120, so we now can find 144/3=48 rather simply by using this method.(6 votes)
- Is there anywhere to practice this on this site?(5 votes)
- Yes - every subject on Khan Academy has practices. Click on "Subjects" in the upper left, and click on a subject, then hit "Practice" and all the practices for the subject will come up - just find the topic you need to work on.
Hope this helps!(6 votes)
- Near1:00, why is h(4) being divided by h(2)? Does 9/4 represent the ratio between h(4) and h(2)? If so, aren't h(4) and h(2) two inputs away from each other. I'm a little confused on how r^2=9/4 and why it works. I get they're representing the same values, but how do we how that that result is the common ratio?(4 votes)
9/4represents the ratio of
h(2)- or more generally the ratio of
h(n+2) / h(n) = (a*r^(n+2)) / a*r^n
h(n+2) / h(n) = (a*r^n*r^2) / a*r^n
h(n+2) / h(n) = r^2
rrepresents our common ratio. Since if we go from
n+1we just multiply by
h(n+1) / h(n) = (a*r^(n+1)) / a*r^n
h(n+1) / h(n) = (a*r^n*r) / a*r^n
h(n+1) / h(n) = r
Maybe reviewing the exponent properties will help you if you're still stuck.(4 votes)
- at1:4281/36 squared so it can also be 9/6 instead of 9/4?(1 vote)
- Not exactly. Be careful about what those fractions equal.
9 / 4 = r ^2 ( which means that r = 3 / 2 )
81 / 36 = r ^2 ( which means that 9 / 6 = r )
But 9/ 6 can be simplified down to 3 / 2 which is the same answer!
Hope this is of help!(6 votes)
- look if anyone who knows this stuff by heart i need your help, im not gonna ask the exact question or even use the numbers just a simple run down so i can understand it better, so in the table kinda like the one in the video only difference is that its and X and Y table, on the X side (left side) it goes down from 1 to 4 but on the Y side it starts at a number and gets bigger as it goes down, my guess is that its growing since the number IS getting bigger, am i right on that? if the number gets bigger on the table its growth but if it starts big and gets smaller overtime then its decay, right? idk, if your still confused i could tell you the exact question since i feel like i worded it weirdly, i also feel like the answer is quite obvious but im not sure.(2 votes)
- the key is not only that it is getting bigger (or smaller) as x goes up. The idea is that it is getting bigger/smaller by the same multiplier every time. So if you start with 1 and the multiplier is 2, then 1*2=2, 2*2=4, 4*2=8, 8*2=16, and it keeps going up faster and faster. In one of the problems on the video, the multiplier was 9/4. So to get smaller, the multiplier has to be less than 1 such as 1/2. Start at 1, 1*1/2=1/4, 1/4 * 1/2=1/8, 1/8*1/2=1/16, etc. Notice that the same pattern is present, one is a whole number and 1 is the denominator of the series.
Is this sort of what you are looking for?(2 votes)
- I solved this in a slightly different way, I just essentially zeroed out the whole equation so that instead of 2 being equal to 144, I made 0 equal to 144.
This way I could substitute 144 in as "a" for 144*r^2=324 which got me r=1.5, then I went all the way back and solved A*1.5^2=144 to get a=64 and r=1.5
My way seemed a lot less complex, is there a downside?(2 votes)
- My logic was that it didn't matter if 2=144 or if 0=144 as long as I maintained the scale as is, by making 0=144 i would change the initial value into 144 instead of an unknown value, then once I have a logically valid initial value I can solve for r which will allow me to find the real initial value of a*r^n=144 once I have r.(2 votes)
- Is there some general formula or rule that can be used to find the common ratio of any exponential equation?
Also couldn't Sal have found "a" by just setting the common ratio equal to h(2)/h(0) ?(1 vote)
- Common ratio divides one tern by the previous term, so you could do h(1)/h(0) and h(2)/h(1) which are next to each other, h(2)/h(0) are two units apart, so that is why it would be (common ratio)^2.(3 votes)
I have a small question. (ps sorry if my english is bad, i'm used to taking math in french...)
At5:35, when 3/2 to the power of n is written, wouldn't there need to be parentheses around the 3/2? with out it wouldn't just the 3 get the to the power of n?
correct me if im wrong please. Have a great day:)(2 votes)
- [Voiceover] Let's say that we have an exponential function, h of n, and since it's an exponential function it's going to be in the form a times r to the n, where a is our initial value and r is our common ratio, and we're going to assume that r is greater than zero. And they've given us some information on h of n. We know that when n is equal to 2, h of 2 is 144, that h of 4 is 324, that h of 6 is 729. So, based on the information here, let's see if we can actually figure out what a and r are going to be. And like always, pause the video and try to give it a go. All right, now let's do this together. So I'm going to focus on r first. The common ratio and if we had successive n's. If we had h of 3, then we could just find the ratio between h of 3 and h of 2 and r would just come out of that, or if the ratio between h of 4 and h of 3, we could solve explicitly for r, but we can get pretty close to that. We can just find the ratio between h of 4 and h of 2. So, h of 4, the ratio between h of 4 and h of 2 is going to be equal to, well, we know h of 4 is 324, and h of 2 is 144, and we could simplify this a little bit. Let's see, if we, if we simplify this, we would get, they're both divisible by 2. If we divide them both by 2, this one in the numerator, 324 divided by 2 is 162. 144 divided by 2 is 72. Let's see, we can divide by 2 again. 81 over 36. Divide by 2, actually, no, we can't divide by 2 anymore, but we can divide by 9 now, so 81 divided by 9 is 9, and 36 divided by 9 is 4. So this thing can be rewritten as 9/4, but we could also rewrite this ratio by using, using this form of an exponential function, so we could also say that this is going to be equal to, h of 4 is going to be a times r. Now n is 4, so r to the 4th power, and h of 2 is going to be a times r to the 2nd power. A times r to the 2nd power, and this simplifies nicely, a divided by a, cancels to just 1, and r to the 4th divided by r-squared, well that's going to be r to the 4 minus 2 power, or r, it's going to be r to the 2nd power, it's going to be r-squared, and so we have a nice little equation set up. R-squared needs to be equal to 9/4. So let me write that down. R-squared is equal to 9/4, and r needs to be greater than zero, so we could just say r is going to be the principal root of 9/4, which is equal to 3/2. So we were able to figure out r. So now how do we figure out, how do we figure out a? Well, there's a couple of ways to do it. You can think about a is going to be what h of zero is equal to, so we could do, I guess we could call it a tabular method, where, let me set up a little table here. So, a little table, so if this is n and this is h of n, so n is zero, we don't know what h of zero is just yet, it's going to be a. We don't know what h of 1 is yet. We do know that h of 2 is 144. And we do know, since the common ratio is 3/2, if we take h of 1 and we multiply it by 3/2, we're going to get h of 2, and if we take h of zero and multiply it by 3/2, we're going to get h of 1. So, h of 1 is going to be 144 divided by 3/2. So let's write that down. H of 1 is equal to 144 divided by 3/2, which is going to be 144, 144 times 2 over 3, and let's see, 144 divided by 3 is going to be equal to, is it, let's see, one 3 goes into, I'll just do this by hand, my brain doesn't work that well while I'm making videos. 3 goes into 14 four times, 4 times 3 is 12, subtract, so it's going to be, I see, it's going to be 48. 3 goes into 24 eight times. 8 times 3 is 24, you have no remainder, so, this is going to be 48 times 2, which is going to be equal to 96 so this is going to be 96, and so if we want to figure out h of zero, we just divide by 3/2 again, so, h of zero is 96 divided by 3/2 which is equal to 96 times 2 over 3. 96 divided by 3, let's see, it's going to be 32, so this is going to be 32 times, did I do that right? Yup, 32 times 2 which is equal to 64. 64. And so, just like that, we've figured out that a is equal to 64, and r, and r is equal to 3/2. So we can write h of n, we can say that h of n is equal to 64, is equal to 64 times 3/2, times 3/2, our common ratio, to the nth power. Now, there's another way that we could have tackled this instead of this tabular method. We could have just solved for a now, since we know r. We know, for example, we know for example that h of 2 which is going to be equal to a times, we know what our common ratio is, it's 3/2 to the 2nd power, is going to be equal to 144. And so, we could say a times 9 over 4 is equal to 144, and so we can multiply both sides times the reciprocal of 9 over 4, so we multiply both sides times 4 over 9, times 4 over 9, and that cancels with that, that cancels with that, and we are left with a is equal to, well let's see, 144 divided by 9 is going to be, is going to be, I want to say it's equal to 16? Is that right? I think that is right. Yup, and then 16 times 4, so this is going to be 16 and 1, 16 times 4 is 64, which is exactly what we got before, and so this actually, this method, now that I look at it, is actually a good bit easier, but they're equivalent, and I like this one because you get to see the common ratio in action, and you get to see that the initial value is the initial value. It is h of zero, but either way, once you figure out r, and you know one of the h of, you know what the function is at a value, you can solve for a, and frankly, if you knew a and you knew what the function was for a given n, you could likewise solve for r. So hopefully you enjoyed that.