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Câu hỏi: Giải phương trình ${{2}^{x+4}}+{{2}^{x+2}}={{5}^{x+1}}+{{4.5}^{x}}.$
A. $x={{\log }_{\dfrac{2}{5}}}\dfrac{9}{20}.$
B. $x={{\log }_{\dfrac{2}{5}}}\dfrac{20}{9}.$
C. $x={{\log }_{\dfrac{5}{2}}}\dfrac{9}{20}.$
D. $x={{\log }_{\dfrac{5}{2}}}\dfrac{20}{9}.$
A. $x={{\log }_{\dfrac{2}{5}}}\dfrac{9}{20}.$
B. $x={{\log }_{\dfrac{2}{5}}}\dfrac{20}{9}.$
C. $x={{\log }_{\dfrac{5}{2}}}\dfrac{9}{20}.$
D. $x={{\log }_{\dfrac{5}{2}}}\dfrac{20}{9}.$
Ta có ${{2}^{x+4}}+{{2}^{x+2}}={{5}^{x+1}}+{{4.5}^{x}}\Leftrightarrow {{2}^{4}}{{.2}^{x}}+{{2}^{2}}{{.2}^{x}}={{5.5}^{x}}+{{4.5}^{x}}\Leftrightarrow {{20.2}^{x}}={{9.5}^{x}}$
$\Leftrightarrow {{\left( \dfrac{5}{2} \right)}^{x}}=\dfrac{20}{9}\Leftrightarrow x={{\log }_{\dfrac{5}{2}}}\dfrac{20}{9}.$
$\Leftrightarrow {{\left( \dfrac{5}{2} \right)}^{x}}=\dfrac{20}{9}\Leftrightarrow x={{\log }_{\dfrac{5}{2}}}\dfrac{20}{9}.$
Đáp án D.