A pandigital number is one which contains every digit from 0 to 9 at least once. 1234567890, 1902837465000000, and 9023289761326634265 are all pandigital.

A diverse pair of numbers is a pair of numbers $(a, b)$ such that $a^b$ is pandigital. $b$ is called the diversifying exponent.

Challenge: Given a number $a$, find the smallest corresponding diversifying exponent $b$. This is a code-golf, so the shortest program in bytes wins.

(You may assume that there exists such an exponent, that is, your program will not be given invalid input, such as a power of 10.)

Your solution must be able to handle at the minimum the given test cases, but it should theoretically handle all valid inputs.

This is A090493 on OEIS.

Test cases

2 -> 68

3 -> 39

4 -> 34

5 -> 19

6 -> 20

7 -> 18

8 -> 28

9 -> 24

11 -> 23

12 -> 22

13 -> 22

14 -> 21

15 -> 12

16 -> 17

17 -> 14

18 -> 21

19 -> 17

20 -> 51

21 -> 17

22 -> 18

23 -> 14

24 -> 19

25 -> 11

26 -> 18

27 -> 13

28 -> 11

29 -> 12

30 -> 39

31 -> 11

32 -> 14

33 -> 16

34 -> 14

35 -> 19

36 -> 10

1234567890 -> 1

JavaScript (Node.js),  51 46  43 bytes

Takes input as a BigInt literal. Returns true instead of 1.

f=(n,k=n)=>new Set(n+'').size>9||1+f(n*k,k)

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J, 25 bytes

>:@]^:(10>#@~.@":@^)^:_&1

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Single monadic verb. The input should be an extended-precision integer (e.g. 2x).

How it works

>:@]^:(10>#@~.@":@^)^:_&1    Monadic verb. Input: base a

    ^:              ^:_      Good old do-while loop.

                       &1    Given 1 as the starting point for b,

>:@]                         increment it each step

      (            )         and continue while the condition is true:

               ":@^          Digits of a^b

            ~.@              Unique digits

          #@                 Count of unique digits

       10>                   is less than 10

Jelly, 12 11 bytes

1*@ṾØDfƑʋ1#

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How it works

1*@ṾØDfƑʋ1#  Main link. Argument: n



1            Set the return value to 1.

         1#  Call the link to the left for k = 1, 2, ... and with right argument n,

             until it returns a truthy value.

        ʋ      Combine the four links to the left into a dyadic chain.

 *@              Compute n**k.

   Ṿ             Convert it to its string representation.

    ØD           Yield "0123456789".

      fƑ         Filter and return 1 is the result is equal to the left argument.

Perl 6, 32 bytes

{first ($_** *).comb.Set>9,1..*}

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Pretty self-explanatory.

Explanation

{                              }  # Anonymous code block

first                     ,1..*   # First positive number that

      ($_** *)    # When the input is raised to that power

              .comb.Set    # The set of digits

                       >9  # Is longer than 9

Python 2, 44 bytes

f=lambda n,k=1:11>len(set(`k`))and-~f(n,n*k)

Input has to be a long, as ​`k`​ behaves differently for longs and ints.

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