Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 09:25, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.3:3h moved to Solution 3.3:3h: Robot: moved page) ← Previous diff		Current revision (07:04, 2 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Because <math>a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}</math>, the logarithm law, <math>b\lg a = \lg a^b</math>, gives that
-	<center> [[Image:3_3_3h.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\log_{a} \bigl(a^{2}\sqrt{a}\,\bigr) = \log_{a}a^{5/2} = \frac{5}{2}\cdot\log_{a}a = \frac{5}{2}\cdot 1 = \frac{5}{2}\,,</math>}}
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		+	where we have used that <math>\log_{a}a = 1\,</math>.
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		+	Note: In this exercise, we assume, implicitly, that <math>a > 0</math> and <math>a\ne 1\,</math>.

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Current revision

Because a2a=a2a12=a2+12=a52 , the logarithm law, blga=lgab, gives that

logaa2a=logaa52=25logaa=251=25

where we have used that logaa=1.

Note: In this exercise, we assume, implicitly, that a0 and a=1.