Convergence of Newton method

Question

I have showed using Taylor's theorem that the error after n iterations $e_n$ satisfies $e_{n+1}=Ce_n^2.$

I am now asked to show that to guarantee an error of less than $10^{-M}$ we need approximately $\log_{10}(M) $ iterations. How would I go about showing this?

Answer 1

$|Ce_{k+1}|\le|Ce_k|^2$ implies $$10^M\sim|Ce_n|\le|Ce_0|^{2^n}$$ which will give with twice applying the logarithm give an a-priori estimate of the step number.