Lösung 3.3:5b

Aus Online Mathematik Brückenkurs 1

By using the logarithm laws,

\displaystyle \lg a+\lg b=\lg \left( a\centerdot b \right)

\displaystyle \text{log }a-\text{ log }b=\text{log}\left( \frac{a}{b} \right)

we can collect together the terms into one logarithmic expression

\displaystyle \begin{align} & \ln 8-\ln 4-\ln 2=\ln 8-\left( \ln 4+\ln 2 \right)=\ln 8-\ln \left( 4\centerdot 2 \right) \\ & =\ln \frac{8}{4\centerdot 2}=\ln 1=0, \\ \end{align}

where ln 1 =0, since \displaystyle e^{0}=1 (the equality \displaystyle a^{0}=1 holds for all \displaystyle a\ne 0 ).