График показательной функции \(f\left( x \right) = {a^{x + b}}\) проходит через точки \(\left( {3;2} \right)\) и \(\left( {5;4} \right)\).
Следовательно: \(\left\{ {\begin{array}{*{20}{c}}{2 = {a^{3 + b}}}\\{4 = {a^{5 + b}}}\end{array}} \right.\)
Разделим второе уравнение на первое: \(\frac{{{a^{5 + b}}}}{{{a^{3 + b}}}} = \frac{4}{2}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{a^2} = 2\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,a = \sqrt 2 .\)
Тогда: \({\sqrt 2 ^{\,3 + b}} = 2\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{\sqrt 2 ^{\,3 + b}} = {\sqrt 2 ^{\,2}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,3 + b = 2\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,b = — 1.\)
Таким образом: \(f\left( x \right) = {\sqrt 2 ^{\,x — 1}}\) и \({\sqrt 2 ^{\,x — 1}} = \frac{1}{8}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,{2^{\frac{{x — 1}}{2}}} = {2^{ — 3}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x = — 5.\)
Ответ: – 5.