Solution 3.3:3h

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Because
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<center> [[Image:3_3_3h.gif]] </center>
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<math>a^{2}\sqrt{a}=a^{2}a^{\frac{1}{2}}=a^{2+\frac{1}{2}}=a^{\frac{5}{2}}</math>, the logarithm law,
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<math>b\lg a=\lg a^{b}</math>, gives that
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<math>\log _{a}a^{2}\sqrt{a}=\log _{a}a^{\frac{5}{2}}=\frac{5}{2}\centerdot \log _{a}a=\frac{5}{2}\centerdot 1=\frac{5}{2},</math>
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where we have used that
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<math>\log _{a}a=1</math>.
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NOTE: In this exercise, we assume, implicitly, that
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<math>\text{a}>0\text{ }</math>
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and
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<math>\text{a}\ne \text{1}</math>.

Revision as of 14:41, 25 September 2008

Because \displaystyle a^{2}\sqrt{a}=a^{2}a^{\frac{1}{2}}=a^{2+\frac{1}{2}}=a^{\frac{5}{2}}, the logarithm law, \displaystyle b\lg a=\lg a^{b}, gives that


\displaystyle \log _{a}a^{2}\sqrt{a}=\log _{a}a^{\frac{5}{2}}=\frac{5}{2}\centerdot \log _{a}a=\frac{5}{2}\centerdot 1=\frac{5}{2},


where we have used that \displaystyle \log _{a}a=1.

NOTE: In this exercise, we assume, implicitly, that \displaystyle \text{a}>0\text{ } and \displaystyle \text{a}\ne \text{1}.

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