From Förberedande kurs i matematik 1

All three arguments of the logarithm can be written as powers of \displaystyle \text{3},

\displaystyle \begin{align} & 27^{\frac{1}{3}}=\left( 3^{3} \right)^{\frac{1}{3}}=3^{3\centerdot \frac{1}{3}}=3^{1}=3, \\ & \frac{1}{9}=\frac{1}{3^{2}}=3^{-2} \\ \end{align}

and it is therefore appropriate to use base \displaystyle \text{3} when simplifying using the logarithms, even if we have the base \displaystyle \text{1}0 logarithm, lg,

\displaystyle \begin{align} & \lg 27^{\frac{1}{3}}+\frac{\lg 3}{2}+\lg \frac{1}{9}=\lg 3+\frac{1}{2}\lg 3+\lg 3^{-2} \\ & =\lg 3+\frac{1}{2}\lg 3+\left( -2 \right)\centerdot \lg 3 \\ & =\left( 1+\frac{1}{2}-2 \right)\lg 3=-\frac{1}{2}\lg 3 \\ \end{align}

This expression cannot be simplified any further.