From Förberedande kurs i matematik 1

Revision as of 09:25, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:3g moved to Solution 3.3:3g: Robot: moved page) ← Previous diff		Current revision (07:00, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Using the logarithm law, <math>\lg a-\lg b = \lg\frac{a}{b}\,</math>, the expression can be calculated as
-	<center> [[Image:3_3_3g.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\log_3 12 - \log_3 4 = \log_3\frac{12}{4} = \log _3 3 = 1\,\textrm{.}</math>}}
		+
		+	Another way is to write <math>12 = 3\cdot 4</math> and use the logarithm law,
		+	<math>\lg (ab) = \lg a + \lg b\,</math>,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\log _{3}12 - \log _{3}4
		+	&= \log_{3}(3\cdot 4) - \log_{3} 4\\[5pt]
		+	&= \log_{3}3 + \log _{3}4 - \log _{3}4\\[5pt]
		+	&= \log _{3}3\\[5pt]
		+	&= 1\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

Using the logarithm law, \displaystyle \lg a-\lg b = \lg\frac{a}{b}\,, the expression can be calculated as

\displaystyle \log_3 12 - \log_3 4 = \log_3\frac{12}{4} = \log _3 3 = 1\,\textrm{.}

Another way is to write \displaystyle 12 = 3\cdot 4 and use the logarithm law, \displaystyle \lg (ab) = \lg a + \lg b\,,

\displaystyle \begin{align} \log _{3}12 - \log _{3}4 &= \log_{3}(3\cdot 4) - \log_{3} 4\\[5pt] &= \log_{3}3 + \log _{3}4 - \log _{3}4\\[5pt] &= \log _{3}3\\[5pt] &= 1\,\textrm{.} \end{align}