From Förberedande kurs i matematik 1

Revision as of 09:39, 2 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_3c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (06:33, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	First, we rewrite the number 0.125 as a fraction which we also simplify
-	<center> [[Bild:3_3_3c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>0\textrm{.}125 = \frac{125}{1000} = \frac{5\cdot 25}{10^3} = \frac{5\cdot 5\cdot 5}{(2\cdot 5)^3} = \frac{1}{2^3} = 2^{-3}\,\textrm{.}</math>}}
		+
		+	Because 0.125 was expressed as a power of 2, the logarithm can be calculated in full
		+
		+	{{Displayed math||<math>\log_2 0\textrm{.}125 = \log_2 2^{-3} = (-3)\cdot\log_2 2 = (-3)\cdot 1 = -3\,\textrm{.}</math>}}

---

Current revision

First, we rewrite the number 0.125 as a fraction which we also simplify

\displaystyle 0\textrm{.}125 = \frac{125}{1000} = \frac{5\cdot 25}{10^3} = \frac{5\cdot 5\cdot 5}{(2\cdot 5)^3} = \frac{1}{2^3} = 2^{-3}\,\textrm{.}

Because 0.125 was expressed as a power of 2, the logarithm can be calculated in full

\displaystyle \log_2 0\textrm{.}125 = \log_2 2^{-3} = (-3)\cdot\log_2 2 = (-3)\cdot 1 = -3\,\textrm{.}