Question

MEDAL! HELP!!
The graph of a power growth, h, passes through the points (2,8) and (4.64).
Determine a formula:

Answer 1

We'll need to assume that the graph is a straight line, because otherwise we don't have enough information.

The graph passes through two points: (2,8) and (4,64). What is the slope of the line passing through these two points? Grafen passerer gennem to punkter: (2,8) og (4,64). Hvad er hældningen på linje gennem disse to punkter?

Answer 2

Once you have this slope, m, choose either point. Substitute the coordinates of this point, along with the slope, into the point-slope form of the equation of a straight line.
Når du har denne hældning, m, vælge enten punkt. Stedfortræder koordinaterne for dette punkt, sammen med hældning ind i point-slope form af ligningen for en lige linje.

Answer 3

I'm am totally lost at the moment,@mathmale.... And how did you know you had to translate to danish, lol.

Answer 4

@mathmale, also that i've had two similar questions before this one, and i read on the net that i had to use the formulae f(x)=b*x^a. And i have to find a by a=log(y2/y1)/log(x2/x1) ... (i don't know how to find this without a calculator...) and then now that i'd have a, i have to find b by using b=y1/x1^a. and then i'll have a and b to find a function using f(x)=b*x^a. - is this right?

Answer 5

I've read your post again and now see the word "power," which to me signifies "exponential growth model." Does this sound familiar or not?
Jeg har læst dit indlæg igen, og nu ser ordet "magt", som for mig betyder "eksponentiel vækstmodel." Lyder det bekendt eller ej?

Answer 6

... yeah, but i can't find anything about the power growth in my textbook that's why i'm confused - but yeah you're right.

Answer 7

Let's assume that this is indeed an exponential function: "i had to use the formulae f(x)=b*x^a."

We could write this as y=b*x^a.

We are given two points, both supposedly on the graph of this exponential (not power) growth function.

Let's substitute them into the equation, one point at a time, to get "two equations in two unknowns:"

y=b*x^a

First point: (2,8)

Answer 8

For this point, x=2 and y=8. Writing out the formula accordingly: y=8=b*(2)^a,
where a and b are unknown.

Second point: (4, 64). Would you write out the same formula in the same way, but with this x=4 and y=64, please?

Answer 9

y = ?? = b*( ?? )^a

Answer 10

yeah... I think . Like 64=b*4^a

Answer 11

When you have these two, can you divide both equations?

Answer 12

Before I answer that, think: What is our goal here?

Answer 13

We have two good equations in two unknowns, a and b. Our goal is to .... ??

Answer 14

find those two?

Answer 15

Yes, to find a and b. Have you solved simultaneous equations such as these before?
What do you think we should do first?

Answer 16

64 = b*4^a
4 = b*2^a ( dividing both? )

Answer 17

I'd suggest a different approach: We must somehow solve one of the equations for either b or a and substitute the resulting equation into the other equation, to eliminate one of the variables.

Answer 18

i did some before you wrote in the beginning, and i got this:

Answer 19

Thank you for sharing that. That approach uses logarithms, which my approach does not. In the end, both approaches will succeed, leading us towards finding values for a and b.

Answer 20

Look at what we now have:

\[8=b*2^a~and~64=b*4^a\]

Answer 21

If you like, we could take the logarithms of both sides of both equations. Want to proceed in that manner? I'd rather do this problem in a way that is familiar to you.

Answer 22

Sorry!

And yeah, we could do that..

Answer 23

the first would be around 0.30 and the second would be around 0.90

Answer 24

Let's try this:

First equation:
8=b(2)^a

Taking the log of both sides,
log 8=log b(2)^a.

Note that b(2)^a is a product: b * 2^a, right?

Answer 25

You could always try checking your answers at any point. If you believe a = .30 and b = .90 (or was it the other way around?), you could subst. those values into both equations and see whether either or both is true.

Answer 26

But I'd really suggest going through the formal solution for a and b, using logs.

What do YOU prefer to do at this point?

Answer 27

I got a to being 3 and b being 1 and the end result being 1x^3... But I don't know if i'm right.
I subtracted log(4) with log(2) and divided it with log(64)-log(8) and that looks like this 0.90/0.30 and that gives me 3 and that's a. And then i got b by isolating b from y=b*x^a and got b=y/x^a being b=8/2^3, which is 1. and if you put that into y=b*x^a --> y=1x^3... right?

Answer 28

In a situation like this, Sulle, I'd prefer you actually try out your equation.

If you found that b = 1 and a =3, then you have y = 1 * x^3; we can test that by substituting (2,4). If you do that, Sulle, will the equation be true or false?

Answer 29

y = 1 * x ^ 3

8 = 1 * 2^3 true or false?

Answer 30

8=1*2^3 . true...

Answer 31

Likewise, test the other point, (4,64):

64 = 1*4^3 True or false?

Answer 32

true

Answer 33

Isn't that GREAT?!?! You've solved this problem largely on your own. Congrats!!

Answer 34

You've demonstrated that you understand this material well. I'm proud of you.

Answer 35

I got confused for a second, lol! but, yay! :D Thanks for the help though! It kinda got me to go the "right" direction! :D And thanks for being proud. I'm proud myself haha.

Answer 36

Looking forward to working with you again! We've "met" before on OS; that's how I learned that Danish is apparently your native language.

Answer 37

Yeah, I remember you helping me out with a question before. Didn't know you'd remember haha lol. But me too! :D You're a great help :)