How to compute huge numbers with python?

Question

I\'m currently trying to find the value of x,

x = (math.log(X) - math.log(math.fabs(p))/math.log(g))

with :

X = 53710695204         


        

Answer 1

Sympy might be interesting for you. You can do symbolic simplifications and adjust the precision you want to use (using mpmath):

import sympy as sy
sy.init_printing() # enable pretty printing in IPython

# Build the expression:
X,p,g = sy.symbols('X,p,g')
expr = (sy.log(X) - sy.log(sy.Abs(p))/sy.log(g))
# expr = expr.simplify()  # doesn't have any benefit in this case

# The values:
vX = 53710695204323513509337733909021562547350740845028323195225592059762435955297110591848019878050853425581981564064692996024279718640577281681757923541806197728862534268310235863990001242041406600195234734872865710114622767319497082014412908147635982838670976889326329911714511434374891326542317244606912177994106645736126820796903212224
vp = 79293686916250308867562846577205340336400039290615139607865873515636529820700152685808430350565795397930362488139681935988728405965018046160143856932183271822052154707966219579166490625165957544852172686883789422725879425460374250873493847078682057057098206096021890926255094441718327491846721928463078710174998090939469826268390010887
vg = 73114111352295288774462814798129374078459933691513097211327217058892903294045760490674069858786617415857709128629468431860886481058309114786300536376329001946020422132220459480052973446624920516819751293995944131953830388015948998083956038870701901293308432733590605162069671909743966331031815478333541613484527212362582446507824584241

# substitute values into variables:
expr2 = expr.subs({X:vX, p:vp,g:vg})

# evaluate to 150 digits with internal precision up to 1000 digits:
print(expr2.evalf(n=150, maxn=1000))

gives as a result:

    769.744342885511619935129482917192487900343653888850271462255718268257261969359878869753342583593581927254506121925469662801405523964742213571689617098

Update: As noted by casevh and David, when using sympy, attention is to be paid to not losing accuracy by using normal floating point numbers as inputs. To clarify, let's calculate 10**log10(10+1e-30), which obviously results in 10+1e-30:

import sympy as sy
import numpy as np

xf = 1e-30

# numpy with floats:
np_x1 = np.log10(10+ xf)
np_yf = 10**np_x1

# sympy with no extra benefit
sy1_x1 = sy.log(10 + xf) / sy.log(10)
sy1_ye = 10**sy1_x1
sy1_yf = sy1_ye.evalf(n=33)

# sympy, done right:
x = sy.symbols('x')
sy2_x1 = sy.log(10 + x) / sy.log(10)
sy2_ye = 10**sy2_x1
sy2_yf = sy2_ye.evalf(n=33, subs={x:xf})

print("correct answer: 10.0000000000000000000000000000010")
print("        numpy:  {:.31f}".format(np_yf))
print("  naive sympy:  " + repr(sy1_yf))
print("correct sympy:  " + repr(sy2_yf))

gives as result:

correct answer: 10.0000000000000000000000000000010
        numpy:  10.0000000000000000000000000000000
  naive sympy:  10.0000000000000017763568394002504
correct sympy:  10.0000000000000000000000000000010