math: handle exponent separately in Log2

This guarantees that powers of two return exact answers.

We could do a multiprecision approximation for the
rest of the answer too, but this seems like it should be
good enough.

Fixes #4567.

R=golang-dev, iant, remyoudompheng
CC=golang-dev
https://golang.org/cl/6943074

src/pkg/math/all_test.go

@@ -2281,6 +2281,13 @@ func TestLog2(t *testing.T) {
 			t.Errorf("Log2(%g) = %g, want %g", vflogSC[i], f, logSC[i])
 		}
 	}
+	for i := -1074; i <= 1023; i++ {
+		f := Ldexp(1, i)
+		l := Log2(f)
+		if l != float64(i) {
+			t.Errorf("Log2(2**%d) = %g, want %d", i, l, i)
+		}
+	}
 }
 
 func TestModf(t *testing.T) {

src/pkg/math/log10.go

@@ -17,5 +17,6 @@ func log10(x float64) float64 {
 func Log2(x float64) float64
 
 func log2(x float64) float64 {
-	return Log(x) * (1 / Ln2)
+	frac, exp := Frexp(x)
+	return Log(frac)*(1/Ln2) + float64(exp)
 }