График логарифмической функции \(f\left( x \right) = {\log _a}\left( {x + b} \right)\) проходит через точки \(\left( { — 1;2} \right)\) и \(\left( { — 4;0} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{2 = {{\log }_a}\left( { — 1 + b} \right)}\\{0 = {{\log }_a}\left( { — 4 + b} \right)}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{ — 1 + b = {a^2}}\\{ — 4 + b = 1}\end{array}} \right. \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{a^2} = b — 1}\\{b = 5\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a = 2.\)
Таким образом: \(f\left( x \right) = {\log _2}\left( {x + 5} \right)\) и \(f\left( {27} \right) = {\log _2}\left( {27 + 5} \right) = 5.\)
Ответ: 5.
Задача 9. На рисунке изображён график функции \(f\left( x \right) = {\log _a}\left( {x + b} \right).\) Найдите \(f\left( {27} \right).\)