From Förberedande kurs i matematik 1

Revision as of 09:35, 2 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_3b.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (06:27, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	Because we are working with <math>\log _{9}</math>, we express 1/3 as a power of 9,
-	<center> [[Bild:3_3_3b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}</math>}}
		+
		+	Using the logarithm laws, we get
		+
		+	{{Displayed math||<math>\log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.}</math>}}

---

Current revision

Because we are working with \displaystyle \log _{9}, we express 1/3 as a power of 9,

\displaystyle \frac{1}{3} = \frac{1}{\sqrt{9}} = \frac{1}{9^{1/2}} = 9^{-1/2}\,\textrm{.}

Using the logarithm laws, we get

\displaystyle \log_9\frac{1}{3} = \log_9 9^{-1/2} = -\frac{1}{2}\cdot\log_9 9 = -\frac{1}{2}\cdot 1 = -\frac{1}{2}\,\textrm{.}