I found this manipulation of the equation for the error bound of the Bisection method here - http://www.maths.uniswa.sz/docs/m311/bisection.pdf
The error bound is $$\frac{b-a}{2^{n+1}} < \epsilon$$
They say "by taking logarithms we obtain" - $$n > \frac{\log(b-a) - \log(2\epsilon)}{\log (2)}$$
But that is not what I got when I maniuplated it. Here's what I did
$$\frac{b-a}{2^{n+1}} < \epsilon$$
$$2^{n+1}>\frac{b-a}{\epsilon}$$
$$2^{n}>\frac{b-a}{2\epsilon}$$
$$\log_2(2^{n})>\log_2(b-a) - \log_2(2\epsilon)$$
$$n>\log_2(b-a) - \log_2(2\epsilon)$$
So is this correct? Why does the answer from that pdf have a denominator of $\log(2)$?