设正实数 $a$，$b$，$c$ 满足 ${{a}^{{{\log }_{3}}7}}=27$，${{b}^{{{\log }_{7}}11}}=49$，${{c}^{{{\log }_{11}}25}}=\sqrt{11}$．求 ${{a}^{{{\left( {{\log }_{3}}7 \right)}^{2}}}}\text{+}{{b}^{{{\left( {{\log }_{7}}11 \right)}^{2}}}}+{{c}^{{{\left( {{\log }_{11}}25 \right)}^{2}}}}$．
【难度】
【出处】
2009年第27届美国数学邀请赛Ⅱ（AIMEⅡ）
【标注】
【答案】
469
【解析】
由对数和指数的性质可知
${{a}^{{{\left( {{\log }_{3}}7\right)}^{2}}}}+{{b}^{{{\left( {{\log }_{7}}11 \right)}^{2}}}}+{{c}^{{{\left({{\log }_{11}}25 \right)}^{2}}}}={{\left( {{a}^{{{\log }_{3}}7}} \right)}^{{{\log}_{3}}7}}+{{\left( {{b}^{{{\log }_{7}}11}} \right)}^{{{\log }_{7}}11}}+{{\left({{c}^{{{\log }_{11}}25}} \right)}^{{{\log }_{11}}25}}$
$\text{=}{{27}^{{{\log }_{3}}7}}+{{49}^{{{\log}_{7}}11}}+{{\left( \sqrt{11} \right)}^{{{\log }_{11}}25}}$
$\text{=}{{3}^{3{{\log }_{3}}7}}+{{7}^{{{\log}_{7}}11}}+{{11}^{\frac{1}{2}{{\log }_{11}}25}}$
$\text{=}{{7}^{3}}+{{11}^{2}}+\sqrt{25}=343+121+5=469$．
${{a}^{{{\left( {{\log }_{3}}7\right)}^{2}}}}+{{b}^{{{\left( {{\log }_{7}}11 \right)}^{2}}}}+{{c}^{{{\left({{\log }_{11}}25 \right)}^{2}}}}={{\left( {{a}^{{{\log }_{3}}7}} \right)}^{{{\log}_{3}}7}}+{{\left( {{b}^{{{\log }_{7}}11}} \right)}^{{{\log }_{7}}11}}+{{\left({{c}^{{{\log }_{11}}25}} \right)}^{{{\log }_{11}}25}}$
$\text{=}{{27}^{{{\log }_{3}}7}}+{{49}^{{{\log}_{7}}11}}+{{\left( \sqrt{11} \right)}^{{{\log }_{11}}25}}$
$\text{=}{{3}^{3{{\log }_{3}}7}}+{{7}^{{{\log}_{7}}11}}+{{11}^{\frac{1}{2}{{\log }_{11}}25}}$
$\text{=}{{7}^{3}}+{{11}^{2}}+\sqrt{25}=343+121+5=469$．
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