Integer Logarithm

Question

Objective

Take \$a \in ℤ_{>1}\$ and \$b \in ℤ_+\$ as inputs. Write a function \$f\$ such that:

$$ f(a,b) = \left\{ \begin{array}{ll} \log_ab & \quad \text{if} \space \log_ab \in ℚ \\ -1 & \quad \text{otherwise} \end{array} \right. $$

Rules

• Types and formats of the inputs doesn't matter.

• Type and format of the output doesn't matter either, but the fraction must be irreducible. In C++, std::pair<int,unsigned> is an example. (Regarding logarithms, the numerator and the denominator need not to be arbitrary-length integers.)

• Regardless of whether the input type can handle arbitrary-length integers, the algorithm must be "as if" it were handling those. So using floating-point numbers is implicitly banned.

• If \$a\$ and \$b\$ aren't in the sets as specified above, the challenge falls in don't care situation.

• \$f\$ may be curried or uncurried.

• As this is a code-golf, the shortest code in bytes wins.

Test cases

a   b   => answer
-----------------
2   8   => 3/1
3   9   => 2/1
8   2   => 1/3
9   3   => 1/2
4   8   => 3/2
64  256 => 4/3
999 1   => 0/1
-----------------
2   7   => -1
10  50  => -1
5   15  => -1

Answer 1

Sledgehammer, 11 bytes

^{Alternatively, 83 bits.}

I ended up forgetting to actually answer this with Sledgehammer for a while...

⣜⢍⡪⡱⣰⡵⠅⣧⣼⣅⣸

Compressed from this Wolfram Language code:

If[NumberQ[x1 = Log[Input[], Input[]]], x1, -1]

, where NumberQ is a built-in that checks whether Log has returned an actual number and not a symbolic expression, i.e. when it is rational.

Answer 2

Mathematica, 47 45 43 29 bytes ^SBCS

If[NumberQ[c=#~Log~#2],c,-1]&

You can try it online!

Thanks to @J42161217 for saving me a total of 16 bytes!!

Answer 3

Python 3, ^{103 101} 97 bytes

Saved 2 bytes thanks to Kevin Cruijssen!!!
Saved 4 bytes thanks to Value Ink!!!

from math import*
def f(a,b):
 l=log(b,a)
 n,m=l.as_integer_ratio()
 return -1if a**n-b**m else l

Try it online!

Answer 4

Charcoal, 88 82 bytes

≔Ｅθ⟦⟧ζ≔²ηＷ⊖⌈θ¿⌊﹪θη≦⊕ηＵＭθ⎇﹪κηκ∧⊞Ｏ§ζλη÷κηＦζＦι⊞υκＵＭυＥζ№λιＵＭυ÷ι⊟Φ⊕⌈ι∧λ¬⌈﹪ιλ¿›⌈υ⌊υ-1Ｉ⊟υ

Try it online! Link is to verbose version of code. Takes input as a list in [value, base] order and outputs numerator and denominator. Edit: Saved 6 bytes by simplifying my check for exclusive factors. Explanation:

≔Ｅθ⟦⟧ζ≔²ηＷ⊖⌈θ¿⌊﹪θη≦⊕ηＵＭθ⎇﹪κηκ∧⊞Ｏ§ζλη÷κη

Factorise both numbers by trial division.

ＦζＦι⊞υκ

Collect the factors from both inputs, in case some factors are only present in one of them.

ＵＭυＥζ№λιＵＭυ÷ι⊟Φ⊕⌈ι∧λ¬⌈﹪ιλ

Divide the multiplicity of each factor in both the base and the value by their GCD.

¿›⌈υ⌊υ-1

If this is not unique then output -1.

Ｉ⊟υ

If it is unique then output it.

Answer 5

05AB1E, 18 17 bytes

-1 byte thanks to Kevin Cruijssen

Ó0ζD€¿÷ʒOĀ}DËiнë®

Try it online!

Ó                   # prime factorize each input
 0ζ                 # zip the factorizations together with filler 0
                    # for each pair of exponents:
    D€¿             #  get the gcd
       ÷            #  divide both by the gcd (reducing the fraction)
        ʒ  }        # filter the pairs, keep only those where:
         O          #  the sum
          Ā         #  is not 0
            DËi     # if all the resulting pairs are equal:
               н    #  output the first pair
                ë   # otherwise:
                 ®  #  output -1

Answer 6

Python 3, 94 79 76 bytes

Nothing clever here; this is the obvious brute force approach.

Edit: Unified return point and switched to lambda syntax.

Edit 2: Shaved off three bytes due to @xnor's next trick and rules clarification.

lambda a,b:next(((i,j)for i in range(b)for j in range(1,a)if a**i==b**j),-1)

Try it online!

Answer 7

bc, 110 bytes

define f(a,b){
c=0
d=1
while(0!=e=a^c-b^d){if(a<2^d){print"-1";return}
if(e>1)d=d+1 else c=c+1}
print c,"/",d}

Try it online!

As with a number of my solutions, it outputs its answer and returns 0, which the test code ignores.

The use of scale= in the provided input is only so the expected value can be passed in. (You can't have a string value in bc.) The code is strictly integer... provided you pass in integers. The code may fail if a or the needed root of a is less than two. (See two failing samples in TIO.)

Answer 8

JavaScript (Node.js),  72  68 bytes

Takes input as (a)(b), where \$a\$ and \$b\$ are BigInts. Returns either [numerator, denominator] or \$-1\$.

a=>g=(b,p=0n,q=1n)=>p>b?-1:(d=a**p-b**q)?g(b,d>0?++q&&p:-~p,q):[p,q]

Try it online!

How?

We start with \$p=0\$ and \$q=1\$. We iteratively increment \$q\$ if \$a^p>b^q\$ or increment \$p\$ if \$a^p<b^q\$, until \$a^p=b^q\$ (success) or \$p>b\$ (no solution).

Answer 9

Ruby, 72 bytes

->a,b{n,d=[*1..a*b].product([*1..a]).find{|n,d|a**n==b**d};n ?n/1r/d:-1}

Try it online!