Solution 3.3:3h
From Förberedande kurs i matematik 1
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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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- | {{ | + | {{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>. |
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\,.