График показательной функции \(f\left( x \right) = {a^x} + b\) проходит через точки \(\left( {0; — 3} \right)\) и \(\left( { — 1; — 2} \right)\). Следовательно:
\(\left\{ {\begin{array}{*{20}{c}}{ — 3 = {a^0} + b}\\{ — 2 = {a^{ — 1}} + b}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = — 4\,\,\,\,\,\,}\\{ — 2 = \frac{1}{a} + b}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\frac{1}{a} = 2\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,a = \frac{1}{2}.\)
Таким образом: \(f\left( x \right) = {\left( {\frac{1}{2}} \right)^x} — 4\) и \({\left( {\frac{1}{2}} \right)^x} — 4 = 12\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,{2^{ — x}} = 16\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,x = — 4.\)
Ответ: – 4.