= Solution
Interpret every logarithm in the empirical profile as base ten with the dimensionless argument $P/(1\ {\rm mbar})$. Put
$$
x=\log_{10}\frac{P}{1\ {\rm mbar}}.
$$
The upper atmosphere has $T=800\ {\rm K}$ for $x\leq0$, while integration of $dT/d\log_{10}P=Ax$ below it gives
$$
T(P)=800\ {\rm K}+\frac A2x^2.
$$
Assume that the stated $1250\ {\rm K}$ <planetary equilibrium temperature> is a reasonable <brightness temperature> at the $1\ {\rm bar}=10^3\ {\rm mbar}$ photosphere. Then
$$
1250=800+\frac A2(3)^2,
\qquad
\boxed{A\simeq100\ {\rm K\,dex^{-2}}}.
$$
At $10\ {\rm bar}$, $x=4$, and therefore
$$
\boxed{T(10\ {\rm bar})\simeq800+50(4)^2=1600\ {\rm K}}.
$$
This estimate neglects day-night variation, wavelength-dependent photospheric pressure, and a possible <radiative-convective boundary>; it treats the retrieved profile as representative of the dayside disk.
\Image[../../../paper-315-pressure-temperature.svg]
{title=Plausible pressure-temperature profile for the hot Jupiter}
{description=The profile is isothermal above one millibar and follows the integrated quadratic logarithmic-pressure law below it, calibrated to 1250 kelvin at one bar.}
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