single tape NTM to single tape DTM equivalence
The claim is that for constant $b$, if $f = O(t(n) b^{t(n)})$ then $f = 2^{O(t(n))}$. Indeed, $xb^x \leq b(b+1)^x$, and so
$$
Ct(n) b^{t(n)} \leq Cb (1+b)^{t(n)} = 2^{\log_2(1+b) \cdot t(n) + \log_2 b + \log_2 C}. 
$$
From here it's not hard to check that $f = O(t(n)b^{t(n)})$ implies $f = 2^{O(t(n))}$.