expression test - How does Mathematica decide that Log[2,8] is integer?

At school we learned that $\log_b(x)=\log(x)/\log(b)$, which is implemented in Mathematica as Log[b, x], but the results are different.

Log[8]/Log[2] // N

returns

3.  (* Real *)

while

Log[2, 8] 

returns

3  (* Integer *)

If I want to validate the argument of my hadamardMatrix[] function as a power of 2:

hadamardMatrix[1] := {{1}}
hadamardMatrix[2] := {{1, 1}, {1, -1}}
hadamardMatrix[n_ /; IntegerQ[Log[2, n]]] := 

       KroneckerProduct[hadamardMatrix[2], hadamardMatrix[n/2]]

How can I be sure that any Log[2, 2^N] will always be regarded as integer?

Answer

An interesting question which I've never specifically considered before.

Some observations:

Log[8]/Log[2] // FullSimplify

Log2[8]


Log[2, 8]

3

3

3

And @@ IntegerQ /@ Log2[2^Range[50000]]


And @@ Table[IntegerQ@Log2[2^RandomInteger[5*^8]], {500}]

True

True

Mathematica documentation explicitly states:

Log2 gives exact integer or rational number results when possible.

Also for Log:

Log gives exact rational number results when possible.

For certain special arguments, Log automatically evaluates to exact values.

I think that based on the combination of the empirical result and the statements in the documentation that it is safe to assume that Log2 will return an integer when given a 2^n number.

As far as how this is determined the Implementation Notes say only:

Log and inverse trigonometric functions use Taylor series and functional relations.

which I'm not sure applies.

---

Timings of Log2 compared to J. M.'s lovely bit-level test:

Do[IntegerQ @ Log2 @ n, {n, 1*^7}] // AbsoluteTiming // First

Do[BitAnd[n, n - 1] == 0, {n, 1*^7}] // AbsoluteTiming // First

15.9120279

6.4116112

And now vectorized:

a = Range@1*^6;

Position[Log2@a, _Integer, {1}] // AbsoluteTiming // First

Position[BitAnd[a, a - 1], 0] // AbsoluteTiming // First

1.4196025

0.0468001