Aus Online Mathematik Brückenkurs 1

By using the logarithm laws,

\displaystyle \begin{align} \ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt] \ln a - \ln b &= \ln\frac{a}{b}\,, \end{align}

we can collect together the terms into one logarithmic expression

\displaystyle \begin{align} \ln 8 - \ln 4 - \ln 2 &= \ln 8 - (\ln 4 + \ln 2)\\[5pt] &= \ln 8 - \ln(4\cdot 2)\\[5pt] &= \ln\frac{8}{4\cdot 2}\\[5pt] &= \ln 1\\[5pt] &= 0\,, \end{align}

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