Aus Online Mathematik Brückenkurs 1

In power form, the expressions become

\displaystyle \begin{align} & \sqrt{2}\left( \sqrt[4]{3} \right)^{3}=2^{{1}/{2}\;}\left( 3^{{1}/{4}\;} \right)^{3}=2^{{1}/{2}\;}3^{{3}/{4}\;}, \\ & \sqrt[3]{2}\centerdot 3=2^{{1}/{3}\;}3^{1} \\ \end{align}

Admittedly, it is true that \displaystyle 2^{{1}/{2}\;}>2^{{1}/{3}\;} and \displaystyle 3^{1}>3^{{3}/{4}\;}, but this does not help us to say anything about how the products are related to each other. Instead, we observe that the exponents \displaystyle \frac{1}{2},\ \ \frac{3}{4},\ \ \frac{1}{3} and \displaystyle \text{1} have \displaystyle \text{3}\centerdot \text{4}=\text{12 } as the lowest common denominator which we can take out:

\displaystyle \begin{align} & 2^{\frac{1}{2}}3^{\frac{3}{4}}=2^{\frac{6}{12}}3^{\frac{3\centerdot 3}{12}}=\left( 2^{6}\centerdot 3^{9} \right)^{\frac{1}{12}}, \\ & 2^{\frac{1}{3}}3^{1}=2^{\frac{4}{12}}3^{\frac{12}{12}}=\left( 2^{4}\centerdot 3^{12} \right)^{\frac{1}{12}}. \\ \end{align}

Now, we can compare the bases \displaystyle \text{2}^{\text{6}}\centerdot \text{3}^{\text{9}} and \displaystyle \text{2}^{\text{4}}\centerdot \text{3}^{\text{12}} with each other and so decide which number is larger.

Because

\displaystyle \frac{\text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}}{\text{2}^{\text{4}}\centerdot \text{3}^{\text{12}}}=2^{6-4}3^{9-12}=2^{2}3^{-3}=\frac{2^{2}}{3^{3}}=\frac{4}{27}<1

the denominator \displaystyle \text{2}^{\text{4}}\centerdot \text{3}^{\text{12}} is larger than the numerator \displaystyle \text{2}^{\text{6}}\centerdot \text{3}^{\text{9}}, which means that \displaystyle \sqrt[3]{2}\centerdot 3 is larger than \displaystyle \sqrt{2}\left( \sqrt[4]{3} \right)^{3}.