b)
$\begin{array}{l}
{2^{x - 3}} = {3^{{x^2} - 5x + 6}} \Leftrightarrow \left( {x - 3} \right)\ln 2 = \left( {{x^2} - 5x + 6} \right)\ln 3\\
\Leftrightarrow \left( {x - 3} \right)\ln 2 = \left( {x - 2} \right)\left( {x - 3} \right)\ln 3\\
\Leftrightarrow \left( {x - 3} \right)\left( {\ln 2 - \left( {x - 2} \right)\ln 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
\ln 2 - \left( {x - 2} \right)\ln 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x - 2 = \frac{{\ln 2}}{{\ln 3}} = {\log _3}2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 2 + {\log _3}2
\end{array} \right.
\end{array}$
