*
   * <p>If the result of this method will be immediately rounded to an {@code int}, {@link
   * #log2(double, RoundingMode)} is faster.
   */
  public static double log2(double x) {
    return log(x) / LN_2; // surprisingly within 1 ulp according to tests
  }

  /**
   * Returns the base 2 logarithm of a double value, rounded with the specified rounding mode to an
   * {@code int}.
   *

    checkPositive("x", x);
    if (fitsInLong(x)) {
      return LongMath.log10(x.longValue(), mode);
    }

    int approxLog10 = (int) (log2(x, FLOOR) * LN_2 / LN_10);
    BigInteger approxPow = BigInteger.TEN.pow(approxLog10);
    int approxCmp = approxPow.compareTo(x);

    /*
     * We adjust approxLog10 and approxPow until they're equal to floor(log10(x)) and

     * `numHashFunctions/bitSize`. So, after n insertions, expected bitCount is `bitSize * (1 - (1 -
     * numHashFunctions/bitSize)^n)`. Solving that for n, and approximating `ln x` as `x - 1` when x
     * is close to 1 (why?), gives the following formula.
     */
    double fractionOfBitsSet = (double) bitCount / bitSize;
    return DoubleMath.roundToLong(

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