assertEquals(BigIntegerMath.log10(valueOf(x), mode), IntMath.log10(x, mode));
      }
    }
  }

  // Relies on the correctness of log10(int, FLOOR) and of pow(int, int).
  @GwtIncompatible // pow()
  public void testLog10Exact() {
    for (int x : POSITIVE_INTEGER_CANDIDATES) {
      int floor = IntMath.log10(x, FLOOR);
      boolean expectSuccess = IntMath.pow(10, floor) == x;

  @GwtIncompatible // TODO
  public void testLog10MatchesBigInteger() {
    for (long x : POSITIVE_LONG_CANDIDATES) {
      for (RoundingMode mode : ALL_SAFE_ROUNDING_MODES) {
        assertEquals(BigIntegerMath.log10(valueOf(x), mode), LongMath.log10(x, mode));
      }
    }
  }

  // Relies on the correctness of log10(long, FLOOR) and of pow(long, int).
  @GwtIncompatible // TODO

        assertEquals(BigIntegerMath.log10(valueOf(x), mode), IntMath.log10(x, mode));
      }
    }
  }

  // Relies on the correctness of log10(int, FLOOR) and of pow(int, int).
  @GwtIncompatible // pow()
  public void testLog10Exact() {
    for (int x : POSITIVE_INTEGER_CANDIDATES) {
      int floor = IntMath.log10(x, FLOOR);
      boolean expectSuccess = IntMath.pow(10, floor) == x;

   *     is not a power of ten
   */
  @GwtIncompatible // TODO
  @SuppressWarnings("fallthrough")
  public static int log10(BigInteger x, RoundingMode mode) {
    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);

     * can narrow the possible floor(log10(x)) values to two. For example, if floor(log2(x)) is 6,
     * then 64 <= x < 128, so floor(log10(x)) is either 1 or 2.
     */
    int y = maxLog10ForLeadingZeros[Long.numberOfLeadingZeros(x)];
    /*
     * y is the higher of the two possible values of floor(log10(x)). If x < 10^y, then we want the

     * can narrow the possible floor(log10(x)) values to two. For example, if floor(log2(x)) is 6,
     * then 64 <= x < 128, so floor(log10(x)) is either 1 or 2.
     */
    int y = maxLog10ForLeadingZeros[Long.numberOfLeadingZeros(x)];
    /*
     * y is the higher of the two possible values of floor(log10(x)). If x < 10^y, then we want the

     * is close to 1 (why?), gives the following formula.
     */
    double fractionOfBitsSet = (double) bitCount / bitSize;
    return DoubleMath.roundToLong(
        -Math.log1p(-fractionOfBitsSet) * bitSize / numHashFunctions, RoundingMode.HALF_UP);
  }

  /** Returns the number of bits in the underlying bit array. */
  @VisibleForTesting
  long bitSize() {
    return bits.bitSize();

        `tf.math.greater`, `tf.math.greater_equal`, `tf.math.igamma`,
        `tf.math.igammac`, `tf.math.invert_permutation`, `tf.math.less`,
        `tf.math.less_equal`, `tf.math.lgamma`, `tf.math.log`, `tf.math.log1p`,
        `tf.math.logical_and`, `tf.math.logical_not`, `tf.math.logical_or`,
        `tf.math.maximum`, `tf.math.minimum`, `tf.math.not_equal`,
        `tf.math.polygamma`, `tf.math.reciprocal`, `tf.math.rint`,

pkg math, const Ln2 = 0.693147  // 693147180559945309417232121458176568075500134360255254120680009/1000000000000000000000000000000000000000000000000000000000000000
pkg math, const Log10E = 0.434294  // 10000000000000000000000000000000000000000000000000000000000000/23025850929940456840179914546843642076011014886287729760333279