Solution 3.3:3d
From Förberedande kurs i matematik 1
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- | { | + | We write the argument of <math>\log_{3}</math> as a power of 3, |
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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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+ | 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>}} |
Current revision
We write the argument of \displaystyle \log_{3} as a power of 3,
\displaystyle 9\cdot 3^{1/3} = 3^2\cdot 3^{1/3} = 3^{2+1/3} = 3^{7/3}\,, |
and then simplify the expression with the logarithm laws
\displaystyle \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{.} |