График логарифмической функции \(f\left( x \right) = b + {\log _a}x\) проходит через точки \(\left( {1; — 2} \right)\) и \(\left( {3; — 1} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 2 = b + {{\log }_a}1}\\{ — 1 = b + {{\log }_a}3}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = — 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{ — 1 = b + {{\log }_a}3}\end{array}} \right.} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\, — 1 = — 2 + {\log _a}3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a = 3.\)
Таким образом: \(f\left( x \right) = — 2 + {\log _3}x\) и \( — 2 + {\log _3}x = 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x = 81.\)
Ответ: 81.