On each iteration of the bisection method the error is halved. So we gain one binary digit of precision on each iteration. I want to find how many decimal digits of precision are gained. So does this look alright -
$E_{k+1} = \frac{1}{2}E_k = (\frac{1}{10})^xE_k$
$\frac{1}{2} = \frac{1}{10^x}$
$2= 10^x$
$x = \log_{10} 2$
$x = 0.30103$
So $0.30103$ decimal digits of precision are gained on each step?