20080712, 03:12  #1 
Oct 2006
2^{2}·5·13 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 kvalue for an arbitrary nvalue? Thanks! Last fiddled with by roger on 20080712 at 03:12 
20080712, 12:34  #2  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,889 Posts 
Quote:
find, for example, the kvalue for an arbitrary nvalue? 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. 

20080713, 04:50  #3  
Oct 2006
2^{2}×5×13 Posts 
Quote:
Quote:
Quote:
Example: if I have the equation y=0.3165*x^{1.947}, by what method do I find the equivilant equation (with a different kvalue) with an nvalue of 2? Could you provide a link I could learn it from if you are unwilling to help? Thank you. 

20080713, 05:22  #4  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Quote:
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 bth 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.). Last fiddled with by cheesehead on 20080713 at 05:46 

20080713, 06:53  #5 
"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) 
20080713, 13:38  #6  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}·1,889 Posts 
Quote:
ask "to find a desired arbitrary nvalue". 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. 

20080713, 13:41  #7  
"Bob Silverman"
Nov 2003
North of Boston
16604_{8} Posts 
Quote:
What do you mean by "equivilant (sic) equation (with a different kvalue) with an nvalue 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? 

20080713, 19:41  #8 
Oct 2006
2^{2}×5×13 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. 
20080713, 22:35  #9 
"Lucan"
Dec 2006
England
194A_{16} Posts 

20080714, 02:43  #10 
Oct 2006
100000100_{2} 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 kvalue with n=2 so that the curve would be the same.
Oh well, guess not. 
20080714, 09:56  #11 
"Brian"
Jul 2007
The Netherlands
2×11×149 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.

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