From Förberedande kurs i matematik 1

Revision as of 09:24, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:3d moved to Solution 3.3:3d: Robot: moved page) ← Previous diff		Current revision (06:36, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	We write the argument of <math>\log_{3}</math> as a power of 3,
-	<center> [[Image:3_3_3d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>9\cdot 3^{1/3} = 3^2\cdot 3^{1/3} = 3^{2+1/3} = 3^{7/3}\,,</math>}}
		+
		+	and then simplify the expression with the logarithm laws
		+
		+	{{Displayed math||<math>\log _3 (9\cdot 3^{1/3}) = \log_3 3^{7/3} = \frac{7}{3}\cdot \log_3 3 = \frac{7}{3}\cdot 1 = \frac{7}{3}\,\textrm{.}</math>}}

---

Current revision

We write the argument of \displaystyle \log_{3} as a power of 3,

\displaystyle 9\cdot 3^{1/3} = 3^2\cdot 3^{1/3} = 3^{2+1/3} = 3^{7/3}\,,

and then simplify the expression with the logarithm laws

\displaystyle \log _3 (9\cdot 3^{1/3}) = \log_3 3^{7/3} = \frac{7}{3}\cdot \log_3 3 = \frac{7}{3}\cdot 1 = \frac{7}{3}\,\textrm{.}