Lösung 3.3:2g

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
We know that
+
We know that <math>10^{\lg x} = x</math>, so therefore we rewrite the exponent as
-
<math>10^{\lg x}=x</math>, so therefore we rewrite the exponent as
+
<math>-\lg 0\textrm{.}1 = (-1)\cdot\lg 0\textrm{.}1 = \lg 0\textrm{.}1^{-1}</math>
-
<math>-\lg 0.1=\left( -1 \right)\centerdot \lg 0.1=\lg 0.1^{-1}</math>
+
by using the log law <math>b\lg a = \lg a^b</math>. This gives
-
by using the log law
+
-
<math>b\lg a=\lg a^{b}</math>. This gives
+
-
 
+
{{Displayed math||<math>10^{-\lg 0\textrm{.}1}=10^{\lg 0\textrm{.}1^{-1}}=0\textrm{.}1^{-1}=\frac{1}{0\textrm{.}1}=10\,\textrm{.}</math>}}
-
<math>10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10</math>
+

Version vom 06:20, 2. Okt. 2008

We know that \displaystyle 10^{\lg x} = x, so therefore we rewrite the exponent as \displaystyle -\lg 0\textrm{.}1 = (-1)\cdot\lg 0\textrm{.}1 = \lg 0\textrm{.}1^{-1} by using the log law \displaystyle b\lg a = \lg a^b. This gives

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:2g“