= Solution
Differentiating the <strong coupling constant> $\alpha_s=g^2/(4\pi)$ gives
$$
\frac{d\alpha_s}{d\ln\mu}=-\frac{\beta_0}{2\pi}\alpha_s^2,\qquad \frac{d}{d\ln\mu}\frac1{\alpha_s}=\frac{\beta_0}{2\pi}.
$$
Integrate this constant derivative and express the integration constant through the <strong-coupling scale> $\Lambda$ at which the one-loop inverse coupling vanishes. The <one-loop running of the strong coupling> is
$$
\boxed{\alpha_s(\mu)=\frac{2\pi}{\beta_0\ln(\mu/\Lambda)}=\frac{4\pi}{\beta_0\ln(\mu^2/\Lambda^2)}.}
$$
The positive-coupling branch has $\mu>\Lambda$ and is perturbatively reliable only when the logarithm is sufficiently large. The pole at $\Lambda$ signals failure of the one-loop expansion, not a prediction of a physically infinite observable. Equivalently, $\Lambda=\mu_0\exp[-2\pi/(\beta_0\alpha_s(\mu_0))]$ is scale-independent to this order; this is <dimensional transmutation>.
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