= Solution
A <renormalized coupling> is defined by a subtraction or reference condition at a <renormalization scale> $\mu$. <Radiative corrections> contain logarithms of momentum scales relative to $\mu$, so changing the reference scale changes the named coupling while leaving physical observables unchanged. Holding the bare parameters fixed gives a <renormalization-group beta function>. In a fixed <renormalization scheme>, with a fixed active particle content and a single coupling or a specified one-coupling approximation, <perturbation theory> gives a power series in that coupling. Its coefficients depend on the fields, representations and scheme; beyond the universal leading coefficients they need not be scheme independent. A theory with several couplings generally has coupled beta functions, so the question's scalar equation is schematic rather than literally the form of every quantum field theory. Also $d/d\ln\mu^2=\tfrac12d/d\ln\mu$, which fixes factors of two.
For the stated leading equation let $t=\ln(\mu^2/\mu_0^2)$. Differentiating the reciprocal gives $d(1/\alpha)/dt=-b_0$, hence
$$
\boxed{\alpha(\mu^2)=\frac{\alpha_0}{1-b_0\alpha_0\ln(\mu^2/\mu_0^2)}.}
$$
For $b_0=-\beta_0<0$, define
$$
\boxed{\Lambda^2=\mu_0^2e^{-1/(\beta_0\alpha_0)}.}
$$
Then $1/\alpha=\beta_0\ln(\mu^2/\Lambda^2)$, giving
$$
\boxed{\alpha(\mu^2)=\frac1{\beta_0\ln(\mu^2/\Lambda^2)},\qquad\mu>\Lambda.}
$$
This is <one-loop running of the strong coupling>, with $\beta_0$ here normalized exactly as in the source equation. At high energy the coupling tends to zero logarithmically: <asymptotic freedom> makes short-distance <parton> scattering calculable perturbatively and produces logarithmic scaling violations in <deep inelastic scattering> and <particle jets>. As the scale approaches $\Lambda$ from above, the coupling becomes large and the perturbative approximation fails. Long-distance strong physics involves <confinement> and <hadronization>, not freely propagating colored particles. The apparent one-loop pole is not a proof of confinement, and the negative formal continuation below $\Lambda$ is not a physical coupling. These are the <infrared limitations of one-loop QCD running>; the https://pdg.lbl.gov/2023/reviews/rpp2022-rev-qcd.pdf[Particle Data Group QCD review] describes the distinction between perturbative running and its nonperturbative scale.
The expected rough <QCD scale> is \b[of order a few hundred MeV], comparable to the inverse hadronic length scale $\hbar c/(1\,\mathrm{fm})\simeq0.2\,\mathrm{GeV}$. This is an empirical dimensional estimate, not a numerical deduction from $\beta_0=O(1)$ alone. Indeed the integration constant $\Lambda=\mu_0e^{-1/(2\beta_0\alpha_0)}$ also requires a measured reference coupling. Its precise value depends on flavor number, perturbative order and renormalization scheme.
For $b_0=+\beta_0$, the same integrated solution instead increases with energy and has its perturbative <Landau pole> at
$$
\boxed{\mu_L^2=\mu_0^2e^{1/(\beta_0\alpha_0)}.}
$$
The coupling decreases toward the infrared in this approximation; toward the ultraviolet, perturbation theory ceases to be reliable before reaching the pole. For the self-coupling-dominated Higgs approximation, use the conventional <Higgs potential> $V=\lambda_H(\phi^\dagger\phi-v^2/2)^2$. Then
$$
m_H^2=2\lambda_Hv^2,\qquad\alpha_H=\lambda_H/(4\pi).
$$
This quartic normalization is separate from Q1's real-triplet normalization. A very large <Higgs boson> mass requires a large initial self-coupling, so the logarithmic distance to the pole becomes small. Requiring a weakly coupled <Standard Model> up to a high scale $M$ demands approximately
$$
\alpha_H(\mu_0)<\frac1{\beta_0\ln(M^2/\mu_0^2)},
$$
with a stronger practical margin to keep the evolved coupling perturbative. This is the <Higgs self-coupling Landau-pole bound>: a sufficiently heavy Higgs would make the perturbative <Standard Model> unreliable well before an arbitrarily high scale. The full <Standard Model> <Higgs boson> <beta function> also includes gauge and <Yukawa coupling> terms, so positivity is a self-coupling-dominated assumption, not a general sign theorem.
For the requested bottom threshold, take the physical color number $N_c=3$. In the source normalization,
$$
\beta_4=\frac{25}{12\pi},\qquad\beta_5=\frac{23}{12\pi}.
$$
Continuity of the leading coupling at $\mu=m_b$ gives
$$
\beta_5\ln\frac{m_b^2}{\Lambda_5^2}=\beta_4\ln\frac{m_b^2}{\Lambda_4^2}.
$$
Divide by $2\beta_5$ to find $\ln(m_b/\Lambda_5)=(25/23)\ln(m_b/\Lambda_4)$. Rearranging and exponentiating yields
$$
\boxed{\Lambda_5=\Lambda_4\left(\frac{m_b}{\Lambda_4}\right)^{-2/23}.}
$$
This is <matching the QCD scale across a quark threshold>. The named integration constant changes when the beta coefficient changes; the matched coupling itself stays continuous at leading order. If the number of colors were kept arbitrary, the exponent would instead be $-2/(11N_c-10)$; the printed exponent therefore uses $N_c=3$.
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