$\log\left(\frac{p\cdot p’}{(E^2-p^2)/2}\right)=\log\left(\frac{-(p’-p)^2}{m^2}\right)$? On page 182 of Peskin & Schroeder, equation (6.17) ends with the equality:
$$2\log\left(\frac{p\cdot p’}{(E^2-p^2)/2}\right)=2\log\left(\frac{-(p’-p)^2}{m^2}\right),\tag{6.17}$$
where $p=E(1,\mathbf v)$ is the four-momentum of a particle with inertial mass $m$ that emits Bremsstrahlung after being accelerated to the four-momentum $p’ = E(1,\mathbf {v’})$. However, it looks like:
$$\begin{aligned}\frac{-(p’-p)^2}{m^2} &= \frac{2p\cdot p’ -2m^2}{m^2} \\&= \frac{p\cdot p’}{m^2/2} - 2 \\&= \frac{p\cdot p’}{(E^2-p^2)/2} - 2\\ 
&< \frac{p\cdot p’}{(E^2-p^2)/2}.\end{aligned}$$
Did I make a mistake, or is this a typo in P&S?
Edit:
Does the following show the limit given as answer by Kilian and addresses Joshua’s comment?
$p = E(1,\mathbf v)$, where $E$ is the energy of the particle, holds only on a special inertial frame. From the mass—energy equivalence, it follows that $E$ must be:
$$E^2 = m^2 + |\mathbf p|^2 = m^2 + E^2|\mathbf v|^2 \implies E^2 = \frac {m^2}{1-|\mathbf v|^2},$$
where $|\mathbf v|\rightarrow 1$ as $|\mathbf p|\rightarrow \infty$. Therefore,
$$\begin{aligned} p\cdot p’ - m^2 &= E^2 - \mathbf p \cdot \mathbf p’ - m^2\\
&= E^2 - m^2 – \mathbf p \cdot \mathbf p’\\
&=  \frac {m^2}{1-|\mathbf v|^2} - \frac {m^2(1-|\mathbf v|^2)}{1-|\mathbf v|^2} - \mathbf p \cdot \mathbf p’\\
&=  \frac {m^2-m^2(1-|\mathbf v|^2)}{1-|\mathbf v|^2} - \mathbf p \cdot \mathbf p’\\
&\rightarrow \frac {m^2}{1-|\mathbf v|^2} - \mathbf p \cdot \mathbf p’ = p\cdot p’.\end{aligned}$$
 A: They take the relativistic limit $\textbf{p}\rightarrow \infty$ and therefore:
$$p\cdot p' - m^2 \approx p\cdot p'$$.
A: You are right. This should not be an equality. It is an approximation which holds when $-q^2\gg m^2$.
Peskin's notation here is really a disaster, in my opinion.
Let's clarify. The argument of the first log should be
$$ \frac{p\cdot p'}{(E^2-|\mathbf p|^2)/2}. $$
The $p^2$ in the denominator is not $p_\mu p^\mu$, although in other places Peskin uses that notation as a shorthand. So in the denominator, we can immediate replace $E^2-|\mathbf p|^2=m^2$.
Next, he says $q^2=(p'-p)^2$. Here he does mean the 4-vector norm, which he sometimes represents with a dot. So that's $$q^2 = q_\mu q^\mu = p_\mu p^\mu + p'_\mu p'^\mu - 2 p_\mu p'^\mu = 2m^2 -2p\cdot p'$$ using $p\cdot p =p'\cdot p'=m^2$.
Note as I mentioned in a comment on another answer that in the relativistic limit $$p\cdot p' = E_p E_{p'}(1-\cos\phi)$$ where $\phi$ is the angle between the 3-vectors $\mathbf p$ and $\mathbf p'$. This can be as small as $0$, corresponding to $\phi=0$. Because $q^2$ can be arbitrarily small, in general you cannot neglect it compared to some fixed number. This has nothing to do with being in the relativistic limit or not. It is because $q$ is a momentum difference. A difference can always be small regardless of the magnitude of the two things you are differencing. (You can see him do this correctly on page 201, above Eq 6.70.)
Now we can rearrange and write $$ 2p\cdot p' = 2m^2-q^2.$$
We see that the condition for neglecting the mass is $-q^2\gg m^2$. Note that this is a stronger condition than just being relativistic. It is only true when $p$ and $p'$ are relativistic and very misaligned with each other. So the last equality should really be a $\approx$.
We can verify later in Peskin's exposition that this is what he really meant. On page 200, Eq 6.64, he estimates a certain integral in the limit $-q^2\to\infty$ and gets something equal to the right hand side of Eq 6.17. Then on the next page, he shows that this integral is proportional to the expression $\mathcal I(\mathbf v,\mathbf v')$, which is what you were looking at in Eq 6.17.
