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Solution 3.3:3d

From Förberedande kurs i matematik 1

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Current revision (06:36, 2 October 2008) (edit) (undo)
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We write the argument of
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We write the argument of <math>\log_{3}</math> as a power of 3,
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<math>\log _{3}</math>
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as a power of
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<math>\text{3}</math>,
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{{Displayed math||<math>9\cdot 3^{1/3} = 3^2\cdot 3^{1/3} = 3^{2+1/3} = 3^{7/3}\,,</math>}}
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<math>9\centerdot 3^{{1}/{3}\;}=3^{2}\centerdot 3^{{1}/{3}\;}=3^{2+\frac{1}{3}}=3^{\frac{7}{3}}</math>
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and then simplify the expression with the logarithm laws
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{{Displayed math||<math>\log _3 (9\cdot 3^{1/3}) = \log_3 3^{7/3} = \frac{7}{3}\cdot \log_3 3 = \frac{7}{3}\cdot 1 = \frac{7}{3}\,\textrm{.}</math>}}
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and then simplify the expression with the logarithm laws:
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<math>\log _{3}\left( 9\centerdot 3^{{1}/{3}\;} \right)=\log _{3}3^{\frac{7}{3}}=\frac{7}{3}\centerdot \log _{3}3=\frac{7}{3}\centerdot 1=\frac{7}{3}.</math>
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Current revision

We write the argument of log3 as a power of 3,

9313=32313=32+13=373 

and then simplify the expression with the logarithm laws

log3(9313)=log3373=37log33=371=37.
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