Hello, I found this formula which makes it possible to make an approximation of the natural logarithm of a number. Formula 35 ln(x)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right); I have deduced the following identities: Formula 36 Limit(2^(n)*((x*y)^(1/2^n)-1)/((x*y)^(1/(2^(n+1)))),n=infinity,right)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right)+Limit(2^(n)*((y)^(1/2^n)-1)/((y)^(1/(2^(n+1)))),n=infinity,right); Formula 37 Limit(2^(n)*((x/y)^(1/2^n)-1)/((x/y)^(1/(2^(n+1)))),n=infinity,right)=Limit(2^(n)*(x^(1/2^n)-1)/(x^(1/(2^(n+1)))),n=infinity,right)-Limit(2^(n)*((y)^(1/2^n)-1)/((y)^(1/(2^(n+1)))),n=infinity,right); Formula 38 Limit(2^(n)*((product(x(i),i=1..infinity))^(1/2^n)-1)/((product(x(i),i=1..infinity))^(1/(2^(n+1)))),n=infinity,right)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity); Formula 39 Limit(2^(n)*((product(x(i)/y(i),i=1..infinity))^(1/2^n)-1)/((product(x(i)/y(i),i=1..infinity))^(1/(2^(n+1)))),n=infinity,right)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity) - sum(Limit(2^(n)*(y(i)^(1/2^n)-1)/(y(i)^(1/(2^(n+1)))),n=infinity,right),i=1..infinity); However, in the numerical results, formula 35 intrigues me about its convergence. It seems that for a given n, we get approximately the same number of exact decimals despite the size of the numbers. by example: X=55 ; u := ln(55) ~  1048576/55*(55^(1/1048576)-1)*55^(2097151/2097152) 4.00733318523247091866270291119 = 4.00733318523490960132292657729 Exact number of decimal places : 11 X=5555555555555 u := ln(5555555555555) ~ 1048576/5555555555555*(5555555555555^(1/1048576)-1)*5555555555555^(2097151/2097152) 29.3458195440203748840441577653 = 29.3458195449780708532896646379 Exact number of decimal places : 9 Moreover, its convergence is not linear with n. Best Regards