Solution 3.3:4c
From Förberedande kurs i matematik 1
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| - | {{ | + | All three arguments of the logarithm can be written as powers of 3, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | 27^{\frac{1}{3}} &= \bigl(3^3\bigr)^{\frac{1}{3}} = 3^{3\cdot\frac{1}{3}} = 3^1 = 3\,,\\[5pt] | ||
| + | \frac{1}{9} &= \frac{1}{3^2} = 3^{-2}\,,\\ | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \lg 27^{\frac{1}{3}} + \frac{\lg 3}{2} + \lg \frac{1}{9} | ||
| + | &= \lg 3 + \frac{1}{2}\lg 3 + \lg 3^{-2}\\[5pt] | ||
| + | &= \lg 3 + \frac{1}{2}\lg 3 + (-2)\cdot\lg 3\\[5pt] | ||
| + | &= \Bigl(1+\frac{1}{2}-2\Bigr)\lg 3\\[5pt] | ||
| + | &= -\frac{1}{2}\lg 3\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | This expression cannot be simplified any further. | ||
Current revision
All three arguments of the logarithm can be written as powers of 3,
| \displaystyle \begin{align}
27^{\frac{1}{3}} &= \bigl(3^3\bigr)^{\frac{1}{3}} = 3^{3\cdot\frac{1}{3}} = 3^1 = 3\,,\\[5pt] \frac{1}{9} &= \frac{1}{3^2} = 3^{-2}\,,\\ \end{align} |
and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg,
| \displaystyle \begin{align}
\lg 27^{\frac{1}{3}} + \frac{\lg 3}{2} + \lg \frac{1}{9} &= \lg 3 + \frac{1}{2}\lg 3 + \lg 3^{-2}\\[5pt] &= \lg 3 + \frac{1}{2}\lg 3 + (-2)\cdot\lg 3\\[5pt] &= \Bigl(1+\frac{1}{2}-2\Bigr)\lg 3\\[5pt] &= -\frac{1}{2}\lg 3\,\textrm{.} \end{align} |
This expression cannot be simplified any further.
