Which numbers are both a power of 3 and a cube?

Which numbers are both a power of 1/3 and a cube root?

@christianp I only consider positive power here...

@christianp Let \(0 < k, p\):

\[

pᵏ = kᵖ

\]\[

k \log p = p \log k

\]\[

\frac {\log p} p = \frac {\log k} k

\]

A trivial solution is when \(k = p\), e.g. when \(p := 3\) and \(p := 1/3\). What about the other?

Let \(f : R₊ → R₊\), \(f(k) := \frac {\log k} k\). It is smooth but not monotonic.

\[

f'(k) = \frac 1 {k²} (1 – \log k) = 0

\]

Here \(k = \exp 1\) is the maximum, and \(f(x) = C\) doesn't have a solution for \(\frac 1 {\exp 1} < C\), has just one for \(C ≤ 0\), and has two for the rest…

@christianp For \(\frac 1 {\exp 1} = C\), there is also just one solution.

When \(p := 1/3\), \(f(p)\) is negative, so the trivial solution is the only.

When \(p := 3\), \(f(p)\) is in the region where there is also the second solution.

Generally, for large \(p > 1\), there must be one large and one small \(k\). Not sure how to express the latter precisely.

@amiloradovsky @christianp

Oh that is fun... You lead me to the Lambert's function... and an approximate solution ... via wolfram...

@kat @christianp Thanks for the reference!

The #Lambert W function is the proper name of what I was looking for:

@amiloradovsky why did you take \(k\) to be the same in both cases?

\(3^6 = 9^3\) is both a power of 3 and a cube.

@christianp

I misunderstood the question.

The generalized problem would then be to find all \(k\) and \(l\) for a fixed \(p\), such that \(0 < k, l, p\) and \(pˡ = kᵖ\). Then

\[

\frac {\log p} p = \frac {\log k} l

\]\[

l = \frac {\log k} P

\]\[

k = \exp (P l)

\]

where \(P := \frac {\log p} p\) and all the snolutions may be obtained by varying either \(k\) or \(l\).

@amiloradovsky @christianp

This does not fully describe the problem as stated. For the first part you have here described "a number that is both a pth power of 3 and the cube of p"; to account for all numbers which are both powers of 3 and cubes required introducing two variables, as \(r = 3ˢ = t³ \).

@kimreece

Yes, @christianp already pointed it out (see above). And, with the two variables, the problem is much simpler, and less interesting.

WRT the notation, I actually used \(p\) to generalize \(3\), the given data, to indicate that it's positive. While for the coefficients to be found — \(k\) and \(l\), to indicate that these don't have to be integers.

However, if I'm not mistaken, e.g. #FORTRAN assumes all the letters from 'i' to 'n' to be integers…

@christianp In which domain are these numbers? If all of R is allowed then it is underdetermined and you get an infinite number of solutions. Do you prefer restriction to Q, or are number fields suitable?

@kimreece @christianp

Number field is the terminology to distinguish the use of the word in algebra and in e.g. agriculture/civil engineering? :)

…or section of vector bundle…

One interesting generalization is: given a type \(A\), say, a finite set, or an algebraic structure, find such types \(X, Y\) that

\[

X→A ≃ A→Y

\]

That is the spaces of functions / maps / homomorphisms are isomorphic.

@kimreece up to you. That's part of what I'm getting at. I don't have a particular answer in mind

erou@erou@mathstodon.xyz@christianp The numbers of the form \(3^{3j}\) for \(j\in \mathbb N\) and the the numbers of the form \(1/3^j\) for \(j\in\mathbb N\) I guess?