Задача 32. На рисунке изображён график функции \(f\left( x \right) = {a^{x + b}}.\) Найдите значение x, при котором \(f\left( x \right) = 81.\)
ОТВЕТ: — 6.
График показательной функции \(f\left( x \right) = {a^{x + b}}\) проходит через точки \(\left( { — 2;9} \right)\) и \(\left( {2;1} \right)\).
Следовательно: \(\left\{ {\begin{array}{*{20}{c}}{9 = {a^{ — 2 + b}}}\\{1 = {a^{2 + b}}}\end{array}} \right.\)
Разделим второе уравнение на первое: \(\frac{{{a^{2 + b}}}}{{{a^{ — 2 + b}}}} = \frac{1}{9}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{a^4} = \frac{1}{9}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,a = \frac{1}{{\sqrt 3 }}.\)
Тогда: \({\left( {\frac{1}{{\sqrt 3 }}} \right)^{2 + b}} = 1\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{\left( {\frac{1}{{\sqrt 3 }}} \right)^{2 + b}} = {\left( {\frac{1}{{\sqrt 3 }}} \right)^0}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,2 + b = 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,b = — 2.\)
Таким образом:
\(f\left( x \right) = {\left( {\frac{1}{{\sqrt 3 }}} \right)^{x — 2}}\) и \({\left( {\frac{1}{{\sqrt 3 }}} \right)^{x — 2}} = 81\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,{3^{ — \,\frac{{x — 2}}{2}}} = {3^4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\, — \,\,\frac{{x — 2}}{2} = 4\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x = — 6.\)
Ответ: – 6.