= Solution
For the <Quantum chromodynamics> <gauge group> $SU(3)$, the coefficients on the two sides of the bottom threshold are $\beta_0^{(4)}=25/3$ and $\beta_0^{(5)}=23/3$. Continuity of the <one-loop running of the strong coupling> at $m_b$ gives
$$
\frac{25}{3}\ln\frac{m_b^2}{\Lambda_4^2}=\frac{23}{3}\ln\frac{m_b^2}{\Lambda_5^2},\qquad \ln\frac{m_b}{\Lambda_5}=\frac{25}{23}\ln\frac{m_b}{\Lambda_4}.
$$
Thus <matching the QCD scale across a quark threshold> yields
$$
\boxed{\Lambda_5=m_b\left(\frac{\Lambda_4}{m_b}\right)^{25/23}=\Lambda_4\left(\frac{m_b}{\Lambda_4}\right)^{-2/23},\qquad p=2,\quad r=23.}
$$
The scale parameter changes because the beta-function coefficient changes, even though the matched coupling is continuous. This is leading-order matching with a fixed flavour number within each region; the stated four-flavour region is the effective region above the charm threshold, not a literal claim that four quarks remain active down to arbitrarily small scales.
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