You can calculate both using series expansions for log x (base 10) or, since more readily available, for ln x (base e) and for e^x respectively.
1) One such possibility for log, or rather ln, is :
ln x = 2{ (x-1)/(x+1) + [(x-1)/(x+1)]^3/3 +[(x-1)/(x+1)]^5/5 + ...}
which converges quite quickly for x near to 1. There are "tricks" to improve accuracy and accelerate convergence for higher x values
Note that ln x = 2.303581.log x
2) For antilog:
x = e^ln x is the relationship between x and ln x. To evaluate this you can use the standard series expansion for e^a (since a=lnx is known and x is sought). The expansion is:
e^a = 1 + a + a^2/2! + a^3/3! +a^4/4! + ....
which again converges quite quickly for small values of a
hope it will help u..
