График логарифмической функции \(f\left( x \right) = {\log _a}\left( {x + b} \right)\) проходит через точки \(\left( {2; — 2} \right)\) и \(\left( { — 1;0} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 2 = {{\log }_a}\left( {2 + b} \right)}\\{0 = {{\log }_a}\left( { — 1 + b} \right)}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{a^{ — 2}} = 2 + b}\\{ — 1 + b = 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\frac{1}{{{a^2}}} = 2 + b}\\{b = 2\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a = \dfrac{1}{2}.\)
Таким образом: \(f\left( x \right) = {\log _{\frac{1}{2}}}\left( {x + 2} \right)\) и \(f\left( {30} \right) = {\log _{\frac{1}{2}}}\left( {30 + 2} \right) = — 5.\)
Ответ: – 5.
Задача 11. На рисунке изображён график функции \(f\left( x \right) = {\log _a}\left( {x + b} \right).\) Найдите \(f\left( {30} \right).\)