Solution 3.3:2g

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.3:2g moved to Solution 3.3:2g: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
We know that
-
<center> [[Image:3_3_2g.gif]] </center>
+
<math>10^{\lg x}=x</math>, so therefore we rewrite the exponent as
-
{{NAVCONTENT_STOP}}
+
<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
 +
 
 +
 
 +
<math>10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10</math>

Revision as of 13:31, 25 September 2008

We know that \displaystyle 10^{\lg x}=x, so therefore we rewrite the exponent as \displaystyle -\lg 0.1=\left( -1 \right)\centerdot \lg 0.1=\lg 0.1^{-1} by using the log law \displaystyle b\lg a=\lg a^{b}. This gives


\displaystyle 10^{-\lg 0.1}=10^{\lg 0.1^{-1}}=0.1^{-1}=\frac{1}{0.1}=10

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_3.3:2g"