# Another table of values question, use inverse proportion and find k?

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• June 9th 2009, 02:08 PM
olen12
Another table of values question, use inverse proportion and find k?
So I have:
10 100
40 6
70 2
100 1

That is inverse square correct?

Now, how would I 'model that using the correct inverse proportion' and find the value of k using 10 100
• June 9th 2009, 03:52 PM
HallsofIvy
Quote:

Originally Posted by olen12
So I have:
10 100
40 6
70 2
100 1

That is inverse square correct?

Now, how would I 'model that using the correct inverse proportion' and find the value of k using 10 100

An inverse proportion is of the form y= k/x or xy= k. If that were correct we would have k= 10(100)= 1000 but 40(6)= 240, not 1000. If it were $y= k/x^2$ or $x^2y= k$ then, depending on which number is x, we have either [tex]k= (10)(100^2)= 100000[tex] or $k= (10^2)(100)= 10000$. In the first case, $(40)(6^2)= 1440$, which is not right, and in the second, $(40^2)(6)= 9600$, also not right. So it is not "inverse square". If you were to try a generic $y^n= k/x^m$ or $x^my^n= k$, we could write $(10^m)(100^n)= k$ and $(40^m)(6^n)= k$. Dividing one equation by the other [tex]\left(\frac{10}{40}\right)^m\left(\frac{100}{6}\ri ght)^n= 1[tex] or $\left(\frac{1}{4}\right)^m\left(\frac{50}{3}\right )^n$. Especially easy is the last pair: $(100^m)(1^n)= 100^m= k$ so that $\left(\frac{100}{10}\right)^m\left(\frac{1}{100}\r ight)^n= 1$. That gives you two equations to solve for m and n.