I have a problem where I need to know what J is. I do x^J and get y. For example, if I do 5^J, I would want to get 55 as y. Same with 4^J = 30. When J is 2.455, it works up to 4 only! I need for decimals of J! Is there an equation or something so I can make a program to calculate J?

I want to see if It is possible to get J to work with everything. For example if J = 2.45568392948... And works with 4^J, 5^J etc. To give the right answers for square pyramid numbers. Like, 4 = 30, 5 = 55 etc. So I don't need a new J value for each.

Let me know if I was unclear! It's a confusing topic :D

  • $\begingroup$ Here is some information about square pyramidal numbers. $\endgroup$ Feb 7, 2016 at 0:31
  • $\begingroup$ Please read this tutorial about how to typeset mathematics on this site. $\endgroup$ Feb 7, 2016 at 0:32
  • $\begingroup$ @N.F.Taussig I read from there. But I'm trying to make a new method of calculating it! I use those formulas to see if what I'm doing is correct. That's why I get 4 to 30 and 5 to 55. $\endgroup$ Feb 7, 2016 at 0:32
  • $\begingroup$ Your edit says you want a formula for the square pyramid numbers. There is one: $n^3/3 + n^2/2 + n/6$ (en.wikipedia.org/wiki/Square_pyramidal_number) but no formula of the form $n^J$, $\endgroup$ Feb 7, 2016 at 0:33
  • $\begingroup$ Note that $\frac{\log 30}{\log 4} \neq \frac{\log 55}{\log 5}$. $\endgroup$ Feb 7, 2016 at 0:34

1 Answer 1


If you know $x,y$ and want $J$, you can take the log of both sides. $$x^J=y\\ J \log x = \log y \\ J=\frac {\log y}{\log x}$$
Use any base of logs that you like. Looking at your examples, though, $5^{2.455} \approx 51.9962, 5^{2.456} \approx 52.0799$, so I don't know how you get $55$. Also $4^{2.455} \gt 30$ but just a bit

Added after the edit about square pyramidal numbers: You will not be able to find a single value of $J$ that works for any set of figurate numbers because they are expressed by a polynomial. The square pyramid numbers are $F(n)=\frac 16n(n+1)(2n+1)$. Any power other than $3$ will grow much faster or much slower than the square pyramid numbers. $J=3$ will be off by the factor $3$ and the smaller terms

  • $\begingroup$ The problem is that I need J to work for all numbers! (If possible). 2.455 won't work for five, but I need enough decimals so it will work for both! And then 6, 7, 8 etc.. $\endgroup$ Feb 7, 2016 at 0:30
  • $\begingroup$ It is not possible, and adding decimals is not the answer. You have the wrong functional form. $\endgroup$ Feb 7, 2016 at 0:33

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