= Solution
Let $\Delta=\theta_1-\theta_0>0$, with $\alpha,\beta_{\max}>0$. <Newton law of cooling> gives the warming and cooling profiles
$$
\theta(t)-\theta_0=\begin{cases}
\Delta(1-e^{-\alpha t}),&0\leq t\leq T,\\
\Delta(1-e^{-\alpha T})e^{-\alpha(t-T)},&T\leq t\leq2T.
\end{cases}
$$
In the stipulated linear model, $\beta(t)=\beta_{\max}(\theta(t)-\theta_0)/\Delta$. The bacterial count obeys $\dot N=-\beta(t)N$, so its logarithmic reduction is the <cumulative thermal destruction under exponential warming and cooling>:
$$
\log\frac{N(0)}{N(2T)}=\int_0^{2T}\beta(t)\,dt.
$$
Writing $q=e^{-\alpha T}$, the warming integral divided by $\beta_{\max}$ is $T-(1-q)/\alpha$ and the cooling contribution is $(1-q)^2/\alpha$. Hence the required implicit equation is
$$
\boxed{\beta_{\max}\left[T-\frac{e^{-\alpha T}(1-e^{-\alpha T})}{\alpha}\right]=20\log10}.
$$
The bracket is zero at zero and has <derivative> $(1-q)(1+2q)>0$ for $T>0$, tending to infinity with $T$. There is therefore one positive solution.
\b[For the hardier species, $T$ is unchanged in this model if $\alpha$ and the achieved $\beta_{\max}$ remain the same.] Raising the oven temperature changes $\Delta$, but its factor cancels against the changed slope of the linear destruction law. The entire normalized temperature history and hence the destruction integral remain the same; the comparison uses both the warming and equal-duration cooling stages.
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