Cho \(a>0,b>0,\) nếu viết \({{\log }_{3}}{{\left( \sqrt[5]{{{a}^{3}}b} \right)}^{\frac{2}{3}}}=\frac{x}{5}{{\log }_{3}}a+\frac{y}{15}{{\log }_{3}}b\) thì \(x+y\) bằng bao nhiêu?

* Hướng dẫn giải

Ta có \({{\log }_{3}}{{\left( \sqrt[5]{{{a}^{3}}b} \right)}^{\frac{2}{3}}}=\frac{2}{3}.\frac{1}{5}{{\log }_{3}}\left( {{a}^{3}}b \right)\)

                                \(=\frac{2}{15}\left( {{\log }_{3}}{{a}^{3}}+{{\log }_{3}}b \right)\)

                                \(=\frac{2}{15}.3.{{\log }_{3}}a+\frac{2}{15}{{\log }_{3}}b\)

                                \(=\frac{2}{5}{{\log }_{3}}a+\frac{2}{15}{{\log }_{3}}b.\)

Vậy \(x=2,y=2\Rightarrow x+y=4.\)