Proving ln(x) using infinite series

[darewinder]

Homework Statement

Well we are given a series of steps done with the number "x" and in the end the end value is ln(x). Basically we are asked to prove why it isn't a coincedience

Homework Equations

I put the steps into an equation, but i can't prove it.

ln(x) =^{lim }_{n->inf} (x^\frac{1} {2^n} -1)*2^n

The Attempt at a Solution

Well plugging in gives me inf times 0 so i thought of solving it using Hopital's rule but i can't get to a form where it is 0/0. I tried factoring, rationalizing but i couldn't get anywhere. I would appreciate if you guys can help me do this little bit.

and I would also appreciate if anyone can show me how to come up with a sequence for this. Second part of the question askes me to come up with my own limit of a sequence to get ln(x). I just flipped changed the sine of - inside the brackets to make it + and added the minus sign to the -2n at the end, but that's like the same thing. So if you guys can give me some ideas it would be great! :)

Thank you

 

[HallsofIvy]

Since n only occurs in 2ⁿ, you can simplify by letting a= 2ⁿ. Then the problem becomes showing that
ln(x)= \lim_{a\rightarrow \infty}(x^(1/a)-1)(a)

Now I would be inclined to "reverse" the function: If y= (x^1/a-1)(a), then x= (y/a+ 1)^a. Do you recognize the limit of that as a common formula for e^y? And if x= e^y, then y= ln(x).

 

[darewinder]

tanks a bunch ill get back on you on the reverse function let me work it out. I am eating right now :) But the limit makes sense (argg why didn't i think of that)

 

[darewinder]

hmm i have a question about this, x= (y/a+ 1)^a.

Wouldn't x=1 when we take the limit as a>>inf?

Thanks for your help

never mind i see how x wouldn't equal to 1 because y has some "a" terms in it so we don't know the ratio. But i don't see how x = e^y. Is the x formula like an equation for the e function?