How do you evaluate #ln 5#?

#ln 5# is the zero of the function #f(x) = e^x-5#

If you can calculate #e^x# then you can calculate approximations for #ln 5# using Newton's method...

#f'(x) = e^x#

Starting with an approximation #a_0#, iterate to get better approximations using the formula:

#a_(i+1) = a_i - (f(x))/(f'(x))=a_i - (e^x-5)/e^x=a_i - 1 + 5/e^x#

As for #e^x# you can use the Maclaurin series, which converges for any value of #x#:

#e^x = sum_(k=0)^oo x^k/(k!)#

I would be interested to know how well this method works using a truncated version of this series for #e^x#. How many terms for #e^x# does it require to get a #5# decimal place approximation for #ln 5#?

I may dig a little deeper...

Using #sum_(k=0)^9 x^k/(k!)# as an approximation for #e^x# and Newton's method, the approximation for #ln 5# I found was:

#ln 5 ~~ 1.6094454239#

Using #sum_(k=0)^10 x^k/(k!)# I found:

#ln 5 ~~ 1.6094389965#

Compare those with the proper value:

#ln 5 ~~ 1.6094379124#

If you do want to evaluate:

#sum_(k=0)^10 x^k/(k!)#

then it may be easier to use the form:

#sum_(k=0)^10 x^k/(k!) = 1+x(1+x/2(1+x/3(1+x/4(1+x/5(1+#

#x/6(1+x/7(1+x/8(1+x/9(1+x/10)))))))))#