bc.digits (100)

function ln (x)
  local x = bc.number (x)  -- argument converted to bignum

  -- if x <= 0.5 or >= 2 take the square root and double the result
  if bc.compare (x, 0.5) ~= 1 or bc.compare (x, 2) == 1 then
    return ln (bc.sqrt (x)) * 2
  end -- if x <= 0.5 or >= 2

  local a = bc.number ((x - 1) / (x + 1))
  local e = bc.number (0)
  local t = bc.number (a)

  for i = 1, 200, 2 do
   local E = e
   
   e = e + (t / i)
   t = t * a * a
    
   -- break if no change
    if e == E then
      break
    end -- if no change

  end -- for 

  return e * 2
end -- function ln

print (ln (100))

Output

4.6051701859880913680359829093687284152022029772575459520666558019351452193547049604719944101791965216

0.5
-1.75
1.0625
-0.87109375
-1.241195679
-0.459433287
-1.788921055
1.20023854
-0.559427448
-1.687040931
0.846107103
-1.284102771
-0.351080073
-1.876742782
1.52216347
0.316981629
-1.899522647
1.608186286
0.58626313
-1.656295543

0.50001
-1.74999
1.062465
-0.871168124
-1.241066099
-0.459754937
-1.788625398
1.199180813
-0.561965379
-1.684194913
0.836512505
-1.300246828
-0.309358186
-1.904297513
1.626349018
0.645011127
-1.583960646
0.508931328
-1.740988903
1.031042361

-- See: http://en.wikipedia.org/wiki/Greatest_common_divisor
-- and  http://en.wikipedia.org/wiki/Euclidean_algorithm

function gcd (a, b)
local old_digits = bc.digits (0)  -- we are using integers
local temp

  a = bc.number (a)
  b = bc.number (b)

  while not bc.iszero (b) do  
     temp = b
     b = bc.mod (a, b)
     a = temp
  end -- while
  
  bc.digits (old_digits)  -- put old decimal places back
  return a
end -- gcd

print (gcd (8, 12))   --> 4
print (gcd (42, 56))  --> 14
print (gcd (206, 40)) --> 2
print (gcd (12070, 41820)) --> 170
print (gcd (10362, 12397)) --> 11

-- See: http://en.wikipedia.org/wiki/Least_common_multiple

function lcm (a, b)
  a = bc.number (a)
  b = bc.number (b)
  return (a * b) / gcd (a, b)
end -- lcm

print (lcm (21, 6))  --> 42
print (lcm (4, 6))   --> 12