Aus Online Mathematik Brückenkurs 1

We write the argument of \displaystyle \log _{3} as a power of \displaystyle \text{3},

\displaystyle 9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}

and then simplify the expression with the logarithm laws:

\displaystyle \log _{3}\left( 9\centerdot 3^{{1}/{3}\;} \right)=\log _{3}3^{\frac{7}{3}}=\frac{7}{3}\centerdot \log _{3}3=\frac{7}{3}\centerdot 1=\frac{7}{3}.