Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 09:50, 2 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_4c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (07:16, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	All three arguments of the logarithm can be written as powers of 3,
-	<center> [[Bild:3_3_4c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	27^{\frac{1}{3}} &= \bigl(3^3\bigr)^{\frac{1}{3}} = 3^{3\cdot\frac{1}{3}} = 3^1 = 3\,,\\[5pt]
		+	\frac{1}{9} &= \frac{1}{3^2} = 3^{-2}\,,\\
		+	\end{align}</math>}}
		+
		+	and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\lg 27^{\frac{1}{3}} + \frac{\lg 3}{2} + \lg \frac{1}{9}
		+	&= \lg 3 + \frac{1}{2}\lg 3 + \lg 3^{-2}\\[5pt]
		+	&= \lg 3 + \frac{1}{2}\lg 3 + (-2)\cdot\lg 3\\[5pt]
		+	&= \Bigl(1+\frac{1}{2}-2\Bigr)\lg 3\\[5pt]
		+	&= -\frac{1}{2}\lg 3\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	This expression cannot be simplified any further.

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Current revision

All three arguments of the logarithm can be written as powers of 3,

273191=3331=3331=31=3=132=3−2

and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg,

lg2731+2lg3+lg91=lg3+21lg3+lg3−2=lg3+21lg3+(−2)lg3=1+21−2lg3=−21lg3.

This expression cannot be simplified any further.