From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	By using the logarithm laws,
-	<center> [[Image:3_3_5b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	\ln a + \ln b &= \ln (a\cdot b)\,,\\[5pt]
		+	\ln a - \ln b &= \ln\frac{a}{b}\,,
		+	\end{align}</math>}}
		+
		+	we can collect together the terms into one logarithmic expression
		+
		+	{{Displayed math||<math>\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}</math>}}
		+
		+	where <math>\ln 1 = 0</math>, since <math>e^{0}=1</math> (the equality <math>a^{0}=1</math> holds for all <math>a\ne 0</math>).

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Current revision

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).