How do you calculate #ln(43)# ?

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Answer 1 · 2017-10-29T17:36:18

# ln(43)#

We can attempt approximating this;

Note that #ln(1+x) = x - x^2/2+ x^3/3 ....# As a standard result of power series.

So hence allowing x = 42

Hence #ln(43) approx 42 - 42^2/2 + 42^3/3 #

But this is out of the domain in what the function is of decent approximation

So it would be more simple to approximate #ln(43/100)#

Answer 2 · 2017-10-29T18:51:51

#ln(43) ~~ 3.761200#

Suppose you know:

#log(2) ~~ 0.30103#

#log(3) ~~ 0.4771212547#

#ln(10) ~~ 2.302585093#

#ln(1+x) = x-x^2/2+x^3/3-x^4/4+...#

Then:

#ln(43) = ln(129/3) = ln(129/128 * 128/3) = ln(129/128 * 2^7/3)#

#=ln(129/128) + 7 ln(2) - ln(3)#

#=ln(1+1/128) + (7 log(2) - log(3)) ln(10)#

Now:

#ln(1+1/128) = 1/128-(1/128)^2/2+(1/128)^3/3-... ~~ 0.00778214#

So:

#ln(43) ~~ 0.00778214+(7 * 0.30103 - 0.4771212547) * 2.302585093 ~~ 3.761200#