Solution 3.3:3h

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Current revision (07:04, 2 October 2008) (edit) (undo)
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Because
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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
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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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{{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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<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 <math>\log_{a}a = 1\,</math>.
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where we have used that
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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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<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>.
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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\,.

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