Fractran, 3 bytes

3/2

Fractran literally only has one builtin, but it happens to do exactly what this task is asking for. (It's also Turing-complete, by itself.)

The language doesn't really have a standardised syntax or interpreter. This interpreter (in a comment to a blog post – it's a very simple language) will accept the syntax shown here. (There are other Fractran interpreters with other syntaxes, e.g. some would write this program as 3 2, or even using 3 and 2 as command line arguments, which would lead to a score of 0+3 bytes. I doubt it's possible to do better than 3 in a pre-existing interpreter, though.)

Explanation

3/2
 /   Replace a factor of
  2  2
3    with 3
     {implicit: repeat until no more replacements are possible}

Python 2, 28 bytes

f=lambda n:n%2*n or 3*f(n/2)

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Recursively divide the number by 2 and multiplies the result by 3, as long as the number is even. Odd numbers return themselves.

32 byte alt:

lambda n:n*(n&-n)**0.58496250072

Try it online. Has some float error. The constant is log_2(3)-1.

Uses (n&-n) to find the greatest power-of-2 factor of n, the converts 3**k to 2**k by raising it to the power of log_2(3)-1.

05AB1E, 4 bytes

Ò1~P

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How it works

Ò     Compute the prime factorization of the input.
 1~   Perform bitwise OR with 1, making the only even prime (2) odd (3).
   P  Take the product of the result.

Haskell, 24 23 bytes

until odd$(*3).(`div`2)

The divide by two and multiply by 3 until odd trick in Haskell.

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Alternative with a lambda instead of a pointfree function and with the same byte count:

odd`until`\x->div(x*3)2

Edit: @ais523 saved a byte in the original version and @Ørjan Johansen one in the alternative version, so both version have still the same length. Thanks!

Brain-Flak, 76 bytes

{{({}[()]<([({})]()<({}{})>)>)}{}([{}]()){{}((({})){}{})<>}<>}<>({}({}){}())

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Explanation

This program works by dividing the number by two and tripling until it gets a remainder of one from the division. Then it stops looping and doubles and adds one to the final number.

More detailed explanation coming soon...

JavaScript (ES6), 19 bytes

f=x=>x%2?x:f(x*1.5)

While the input is divisible by two, multiplies it by 1.5, which is equivalent to dividing by 2 and multiplying by 3.

MATL, 7, 6 bytes

Yf1Z|p

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1 byte saved thanks to Dennis's genius observation

The best way to explain this is to show the stack at various points.

Yf  % Prime factors

[2 2 2 3]

1Z| % Bitwise OR with 1

[3 3 3 3]

p   % Product

81

Alternate solution:

Yfto~+p

05AB1E, 6 5 bytes

Saved a byte thanks to Adnan.

ÒDÈ+P

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Explanation

Ò       # push list of prime factors of input
 D      # duplicate
  È     # check each factor for evenness (1 if true, else 0)
   +    # add list of factors and list of comparison results
    P   # product

Alice, 9 bytes

2/S 3
o@i

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Alice has a built-in to replace a divisor of a number with another. I didn't think I'd actually get to make use of it so soon...

Using the code points of characters for I/O, this becomes 6 bytes: I23SO@.

Explanation

2   Push 2 (irrelevant).
/   Reflect to SE. Switch to Ordinal.
i   Read all input as a string.
    The IP bounces up and down, hits the bottom right corner and turns around,
    bounces down again.
i   Try to read more input, but we're at EOF so this pushes an empty string.
/   Reflect to W. Switch to Cardinal.
2   Push 2.
    The IP wraps around to the last column.
3   Push 3.
S   Implicitly discard the empty string and convert the input string to the integer
    value it contains. Then replace the divisor 2 with the divisor 3 in the input.
    This works by multiplying the value by 3/2 as long as it's divisible by 2.
/   Reflect to NW. Switch to Ordinal.
    Immediately bounce off the top boundary. Move SW.   
o   Implicitly convert the result to a string and print it.
    Bounce off the bottom left corner. Move NE.
/   Reflect to S. Switch to Cardinal.
@   Terminate the program.

Mathematica, 22 19 bytes

Thanks to lanlock4 for saving 3 bytes!

#//.x_?EvenQ:>3x/2&

Pure function that does the replacement repeatedly, one factor of 2 at a time. Works on all positive integers less than 2⁶⁵⁵³⁷.

Java, 38 bytes

int f(int n){return n%2>0?n:f(n/2*3);}

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Previous 43-byte solution:

int f(int n){for(;n%2<1;)n=n/2*3;return n;}

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Retina, 23 bytes

.+
$*
+`^(1+)\1$
$&$1
1

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Jelly, 8 5 bytes

Æf3»P

Æf3»P  Main Link, argument is z
Æf     Prime factors
  3»   Takes maximum of 3 and the value for each value in the array
    P  Takes the product of the whole thing

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-3 bytes thanks to a hint from @Dennis!

Pyth - 14 10 9 bytes

*^1.5/PQ2

Counts the number of 2s in the prime factorization (/PQ2). Multiplies the input by 1.5^(# of 2s)

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Brachylog, 7 bytes

~×₂×₃↰|

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How it works

~×₂×₃↰|      original program
?~×₂×₃↰.|?.  with implicit input (?) and output (.) added

?~×₂         input "un-multiplied" by 2
    ×₃       multiplied by 3
      ↰      recursion
       .     is the output
        |    or (in case the above fails, meaning that the input
                 cannot be "un-multiplied" by 2)
         ?.  the input is the output

J, 11 bytes

[:*/q:+2=q:

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[: cap (placeholder to call the next verb monadically)

*/ the product of

q: the prime factors

+ plus (i.e. with one added where)

2 two

= is equal to (each of)

q: the prime factors

J, 15 12 10 bytes

(+2&=)&.q:

Try it online! Works similar to below, just has different logic concerning replacement of 2 with 3.

15 bytes

(2&={,&3)"+&.q:

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Explanation

(2&={,&3)"+&.q:
           &.    "under"; applies the verb on the right, then the left,
                 then the inverse of the right
             q:  prime factors
(       )"+      apply inside on each factor
     ,&3         pair with three: [factor, 3]
 2&=             equality with two: factor == 2
    {            list selection: [factor, 3][factor == 2]
                 gives us 3 for 2 and factor for anything else
           &.q:  under prime factor

Pyth, 9 bytes

Integer output \o/

*F+1.|R1P

Test suite.

How it works

*F+1.|R1P
        P   prime factorization
    .|R1    bitwise OR each element with 1
*F+1        product

Japt, 19 16 10 9 7 bytes

k ®w3Ã×

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Explanation

 k ®   w3Ã ×
Uk mZ{Zw3} r*1
U              # (input)
 k m           # for every prime factor
    Z{Zw3}     # replace it by the maximum of itself and 3
           r*1 # output the product

Japt, 7 bytes

k mw3 ×

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Explanation

k mw3 ×

k        // Factorize the input.
  mw3    // Map each item X by taking max(X, 3).
      ×  // Take the product of the resulting array.
         // Implicit: output result of last expression

CJam, 10 bytes

rimf2Zer:*

Really simple.

Explanation:

ri  e# Read integer:  | 28
mf  e# Prime factors: | [2 2 7]
2   e# Push 2:        | [2 2 7] 2
Z   e# Push 3:        | [2 2 7] 2 3
er  e# Replace:       | [3 3 7]
:*  e# Product:       | 63

APL (Dyalog Unicode), 26 bytes

{⍵(⊣×(3*⊢)×(÷2*⊢))2⍟⍵∨2*⍵}

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This is too verbose, I must be doing something wrong...

R, 42 bytes

The only right amount of bytes in an answer.

x=gmp::factorize(scan());x[x==2]=3;prod(x)

Pretty straightforward, uses the gmp package to factorize x, replaces 2s by 3s, and returns the product.

PHP, 36 Bytes

for($a=$argn;$a%2<1;)$a*=3/2;echo$a;

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