First, we can start with the general definition of the Taylor series expansion, which is:

#sum_(n=0)^N f^((n))(a)/(n!)(x-a)^n#

where #f^((n))(a)# is the #n#th derivative of #f(x)# evaluated at #x -> a#, #n# varies, #a# does not, and #n!# is #1xx2xx3xxcdotsxx(n-1)xxn#. Note that #x->a# does not apply to #(x-a)^n# (yes, I've seen that happen).

Thus you have to take the derivative up to some #n#th order term. Let's say #n = 4#. Then you have:

#f^((0))(x) = f(x) = lnx#

#f'(x) = 1/x#

#f''(x) = -1/x^2#

#f'''(x) = 2/(x^3)#

#f''''(x) = -6/(x^4)#

Now you have the final result to simplify:

#sum_(n=0)^(4) = (f(2))/(0!)(x-2)^0 + (f'(2))/(1!)(x-2)^1 + (f''(2))/(2!)(x-2)^2 + (f'''(2))/(3!)(x-2)^3 + (f''''(2))/(4!)(x-2)^4 + ...#

#= (ln2) + (1/2)(x-2) + (-1/4)/(2)(x-2)^2 + (2/8)/(6)(x-2)^3 + (-6/16)/(24)(x-2)^4 + ...#

#= color(blue)(ln2 + 1/2(x-2) - 1/8(x-2)^2 + 1/24(x-2)^3 - 1/64(x-2)^4 + ...)#

Just remember the following:

- #x->a# only for #f(a)#, not #(x-a)^n#

- #n# varies, but #a# does not

- It's probably wise to take the #n#th derivatives first so you aren't doing them as you are writing out the next step from the point you write out the general series formula