Manipulating powers

roger · 2008-07-12, 03:12

Hi, sorry if this has a really obvious answer

How do you manipulate the power and coefficient in the equation y=kx^n to find, for example, the k-value for an arbitrary n-value?

Thanks!

R.D. Silverman · 2008-07-12, 12:34

Quote:

Originally Posted by roger

Hi, sorry if this has a really obvious answer

How do you manipulate the power and coefficient in the equation y=kx^n to find, for example, the k-value for an arbitrary n-value?

Thanks!

Gibberish. What do you mean by

find, for example, the k-value for an arbitrary n-value?

And the word 'manipulate' has no mathematical meaning.

Solving the equation for either n or k as a function of the other variables
is 1st/2nd year high school algebra. If indeed you do not understand
even this level of mathematics then you need to go learn it before
participating further in this group.

roger · 2008-07-13, 04:50

Quote:

Solving the equation for either n or k as a function of the other variables is 1st/2nd year high school algebra.

I know it should be, but it still hasn't been taught at my school.

Quote:

If indeed you do not understand even this level of mathematics then you need to go learn it before participating further in this group.

I'm here to learn, not to participate by giving mathematical insights. I know I'm not anywhere near that level, and I'm not trying to pass myself off as being there.

Quote:

Gibberish. What do you mean by find, for example, the k-value for an arbitrary n-value?

I mean that, given a 'k' and 'n' value in the form y=kxⁿ, how do you use those numbers to find a desired arbitrary n-value?

Example: if I have the equation y=0.3165*x^1.947, by what method do I find the equivilant equation (with a different k-value) with an n-value of 2?

Could you provide a link I could learn it from if you are unwilling to help?

Thank you.

cheesehead · 2008-07-13, 05:22

Quote:

Originally Posted by roger

Example: if I have the equation y=0.3165*x^1.947, by what method do I find the equivilant equation (with a different k-value) with an n-value of 2?

What may bother some folks reading this is that there's more than one way to interpret your questions (both the original general question and this later one with specific values), and those folks don't like to treat such questions as a sort of multivalued function or relation with more than one possible answer. (Might contemplation of the CASE statement in certain programming languages be helpful to some of those folks? Probably that's not the most useful suggestion.) To be fair, OTOH, it does take more time to compose an inclusive response to such a question, and some folks are genuinely too busy to bother doing such tedium, but in that case, those busy folks really should leave these "multivalued" questions to be answered by those of us with more time.

roger, if you mean that you want to choose values of a and b such that a*x^b = 0.3165*x^1.947 for any x, the only solution is a=0.3165, b=1.947. If the exponent of x (b) has any other value, the shape of the two curves will be different, and there's no way to make them match by changing the multiplying constant (a).

If you mean that for a particular y value, say 4.5 (so that 0.3165*x^1.947 = 4.5), you want to solve for the x value that will make that equation true, first divide both sides by 0.3165, then find the 1/1.947 power of each side. (So, x = (4.5/0.3165)^1/1.947.)

If you mean that for a particular y value, say 9.8 (so that 0.3165*x^1.947 = 9.8), you want to find other a and b values so that a*x^b = 9.8 for some (not every) x, you can choose any nonzero a and b you want, because then you just divide 9.8 by a and find the b-th root (1/b power) to get the x value corresponding to y = 9.8 with those particular a and b values.

If you mean something else, try stating what you want a different way, and give us an example with specific values to show us what you mean. (Note how useful it was to me to be able to incorporate your specific values from your second posting.).

davieddy · 2008-07-13, 06:53

Given two pairs (x1,y1) and (x2,y2) of x,y values (>0) you
can determine the unique values of k and n.

It may help to use

log(y) = log(k) + n*log(x)

R.D. Silverman · 2008-07-13, 13:38

Quote:

Originally Posted by roger

I know it should be, but it still hasn't been taught at my school.

I'm here to learn, not to participate by giving mathematical insights. I know I'm not anywhere near that level, and I'm not trying to pass myself off as being there.

I mean that, given a 'k' and 'n' value in the form y=kxⁿ, how do you use those numbers to find a desired arbitrary n-value?

.

This is nonsense. First you say "given a 'k' and 'n' value " and then you
ask "to find a desired arbitrary n-value".

If you are given n, then you don't need to "find" it!!!!!!!

You need to state what you are trying to do. And I don't buy the
claim that "your school didn't teach it". This is basic level algebra.

R.D. Silverman · 2008-07-13, 13:41

Quote:

Originally Posted by roger

Example: if I have the equation y=0.3165*x^1.947, by what method do I find the equivilant equation (with a different k-value) with an n-value of 2?

.

Another followup...

What do you mean by "equivilant (sic) equation (with a different k-value) with an n-value of 2?" What is an "equivalent" equation? The only function
that is everywhere equal to 0.3165*x^1.947, is
0.3165*x^1.947, . What makes you think there is another?

roger · 2008-07-13, 19:41

Thanks cheesehead, I meant the first option you gave. I'm surprised this isn't possible. Could you provide a topic name or link so I could read about it?

@Silverman: I didn't say that my high school doesn't teach it, I just said I haven't been taught it yet.

davieddy · 2008-07-13, 22:35

Quote:

Originally Posted by roger

Thanks cheesehead, I meant the first option you gave. I'm surprised this isn't possible. Could you provide a topic name or link so I could read about it?

How could you turn a parabola into a straight line by scaling?

roger · 2008-07-14, 02:43

I didn't want to turn it into a straight line, I just wanted to change the exponent to a whole number, and I thought that I could solve for the k-value with n=2 so that the curve would be the same.

Oh well, guess not.

Brian-E · 2008-07-14, 09:56

It often helps to try out very simple examples to get a feel of the situation if the problem lends itself to do that, and this one certainly does. In your equation y=kx^n put k=1 and n=2, and then write a table of values of y for the integer values of x (integers for ease of calculation). You have the square numbers. Now change n from 2 to 3 and do the same: you get the cubes. Simply look at how much faster the cubes increase than the squares (for x>1), and you can easily see that there is no way of changing the multiplying constant k to make the results y the same.

2008-07-12, 03:12	#1
roger Oct 2006 260₁₀ Posts	Manipulating powers Hi, sorry if this has a really obvious answer How do you manipulate the power and coefficient in the equation y=kx^n to find, for example, the k-value for an arbitrary n-value? Thanks! Last fiddled with by roger on 2008-07-12 at 03:12

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2008-07-13, 06:53	#5
davieddy "Lucan" Dec 2006 England 2·3·13·83 Posts	Given two pairs (x1,y1) and (x2,y2) of x,y values (>0) you can determine the unique values of k and n. It may help to use log(y) = log(k) + n*log(x)

2008-07-13, 19:41	#8
roger Oct 2006 404₈ Posts	Thanks cheesehead, I meant the first option you gave. I'm surprised this isn't possible. Could you provide a topic name or link so I could read about it? @Silverman: I didn't say that my high school doesn't teach it, I just said I haven't been taught it yet.

2008-07-14, 02:43	#10
roger Oct 2006 2²×5×13 Posts	I didn't want to turn it into a straight line, I just wanted to change the exponent to a whole number, and I thought that I could solve for the k-value with n=2 so that the curve would be the same. Oh well, guess not.

2008-07-14, 09:56	#11
Brian-E "Brian" Jul 2007 The Netherlands 3279₁₀ Posts	It often helps to try out very simple examples to get a feel of the situation if the problem lends itself to do that, and this one certainly does. In your equation y=kx^n put k=1 and n=2, and then write a table of values of y for the integer values of x (integers for ease of calculation). You have the square numbers. Now change n from 2 to 3 and do the same: you get the cubes. Simply look at how much faster the cubes increase than the squares (for x>1), and you can easily see that there is no way of changing the multiplying constant k to make the results y the same.