График логарифмической функции \(f\left( x \right) = b + {\log _a}x\) проходит через точки \(\left( {1; — 1} \right)\) и \(\left( {2; — 3} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 1 = b + {{\log }_a}1\,}\\{ — 3 = b + {{\log }_a}2}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = — 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{ — 3 = b + {{\log }_a}2}\end{array}} \right.} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\, — 3 = — 1 + {\log _a}2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a = \frac{1}{{\sqrt 2 }}.\)
Таким образом: \(f\left( x \right) = — 1 + {\log _{_{\frac{1}{{\sqrt 2 }}}}}x\) и \(f\left( {\frac{1}{8}} \right) = — 1 + {\log _{_{\frac{1}{{\sqrt 2 }}}}}\frac{1}{8} = — 1 + 6 = 5.\)
Ответ: 5.
Задача 4. На рисунке изображён график функции \(f\left( x \right) = b + {\log _a}x.\) Найдите \(f\left( {0,125} \right).\)