i remember doing something like this last year, but it was just glanced over, and hardly mentioned this year at all. it seems like the kind of thing thats easy once you figure out what to do, but figuringthat out is the whole battle. so, not nly do i not know what to do, i don't even know how to begin to show my work

anyways we have:

Use the following theorem with x=1/3 and m=5 to calculate approximations to log 2. retain 9 decimal places and optain the inequalities .6931460<x<.6931476

and ill just put the whole theorem because its a doozy and i cant make heads or tails of it:

If 0<x<1 and m>_1 we have

log((1+x)/(1-x))=2(x+x^3/3 + .... + (x^ (2m-1))/(2m-1)+ R(subscript m) (x)

where the error term, R(subscript m)(x) satisfies the inequalities

(x^(2m+1))/(2m+1)< R (subscript m)(x)< [(2-x)/(1-x)]*[(x^(2m-1))/(2m+1)]

so, as you can see thats quite a mouthfull...i am pretty sure i copied it down perfectly, but im sure it would be a plus if you are familiar with that theorem, as i had to use so many parenthesis, it must be very hard to understand just by looking at it

thanks in advance for any help with solving this