График логарифмической функции \(f\left( x \right) = b + {\log _a}x\) проходит через точки \(\left( {1; — 3} \right)\) и \(\left( {2; — 2} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 3 = b + {{\log }_a}1}\\{ — 2 = b + {{\log }_a}2}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = — 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{ — 2 = b + {{\log }_a}2}\end{array}} \right.} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\, — 2 = — 3 + {\log _a}2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a = 2.\)
Таким образом: \(f\left( x \right) = — 3 + {\log _2}x\) и \( — 3 + {\log _2}x = 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x = 64.\)
Ответ: 64.