График логарифмической функции \(f\left( x \right) = {\log _a}\left( {x + b} \right)\) проходит через точки \(\left( { — 1; — 1} \right)\) и \(\left( { — 3;0} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 1 = {{\log }_a}\left( { — 1 + b} \right)}\\{0 = {{\log }_a}\left( { — 3 + b} \right)}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\frac{1}{a} = — 1 + b}\\{ — 3 + b = 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a = \frac{1}{{b — 1}}}\\{b = 4\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a = \frac{1}{3}.\)
Таким образом: \(f\left( x \right) = {\log _{\frac{1}{3}}}\left( {x + 4} \right)\) и \(f\left( {77} \right) = {\log _{\frac{1}{3}}}\left( {77 + 4} \right) = — 4.\)
Ответ: – 4.
Задача 12. На рисунке изображён график функции \(f\left( x \right) = {\log _a}\left( {x + b} \right).\) Найдите \(f\left( {77} \right).\)