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
).
