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# Analyzing tables of exponential functions

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.

## Want to join the conversation?

• how does he divide 144 by 3 IN HIS HEAD?
• 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.
• Near , 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?
• Yes `9/4` represents the ratio of `h(4)` and `h(2)` - or more generally the ratio of `h(n+2)` and `h(n)`.

``h(n+2) / h(n) = (a*r^(n+2)) / a*r^nh(n+2) / h(n) = (a*r^n*r^2) / a*r^nh(n+2) / h(n) = r^2``

`r` represents our common ratio. Since if we go from `n` to `n+1` we just multiply by `r` once more.

``h(n+1) / h(n) = (a*r^(n+1)) / a*r^nh(n+1) / h(n) = (a*r^n*r) / a*r^nh(n+1) / h(n) = r``

Maybe reviewing the exponent properties will help you if you're still stuck.
• Is there anywhere to practice this on this site?
• 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!
• am i the only one who did a system of equation ?

144 = a*r²
324 = a*r^4

So a = 144/r²
and 324 = 144/r² * r^4 then 324=144r², so 324/144=r²

rest is just the result of sal
• at 81/36 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!
• dude, whats even going on? you lost me at the beginning
• 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.
• 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?
• 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?