Lösung 3.3:4c
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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All three arguments of the logarithm can be written as powers of 3, | All three arguments of the logarithm can be written as powers of 3, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
27^{\frac{1}{3}} &= \bigl(3^3\bigr)^{\frac{1}{3}} = 3^{3\cdot\frac{1}{3}} = 3^1 = 3\,,\\[5pt] | 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}\,,\\ | \frac{1}{9} &= \frac{1}{3^2} = 3^{-2}\,,\\ | ||
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and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg, | and it is therefore appropriate to use base 3 when simplifying using the logarithms, even if we have the base 10-logarithm, lg, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\lg 27^{\frac{1}{3}} + \frac{\lg 3}{2} + \lg \frac{1}{9} | \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 + \lg 3^{-2}\\[5pt] | ||
Version vom 08:44, 22. Okt. 2008
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.
