From Förberedande kurs i matematik 1

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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> [[Bild: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 \displaystyle a^{2}\sqrt{a} = a^{2}a^{1/2} = a^{2+1/2} = a^{5/2}, the logarithm law, \displaystyle b\lg a = \lg a^b, gives that

\displaystyle \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}\,,

where we have used that \displaystyle \log_{a}a = 1\,.

Note: In this exercise, we assume, implicitly, that \displaystyle a > 0 and \displaystyle a\ne 1\,.