Solution 3.3:4c

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.3:4c moved to Solution 3.3:4c: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
All three arguments of the logarithm can be written as powers of
-
<center> [[Image:3_3_4c.gif]] </center>
+
<math>\text{3}</math>,
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\begin{align}
 +
& 27^{\frac{1}{3}}=\left( 3^{3} \right)^{\frac{1}{3}}=3^{3\centerdot \frac{1}{3}}=3^{1}=3, \\
 +
& \frac{1}{9}=\frac{1}{3^{2}}=3^{-2} \\
 +
\end{align}</math>
 +
 
 +
 
 +
 
 +
and it is therefore appropriate to use base
 +
<math>\text{3}</math>
 +
when simplifying using the logarithms, even if we have the base
 +
<math>\text{1}0</math>
 +
logarithm, lg,
 +
 
 +
 
 +
<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} \\
 +
& =\lg 3+\frac{1}{2}\lg 3+\left( -2 \right)\centerdot \lg 3 \\
 +
& =\left( 1+\frac{1}{2}-2 \right)\lg 3=-\frac{1}{2}\lg 3 \\
 +
\end{align}</math>
 +
 
 +
 
 +
This expression cannot be simplified any further.

Revision as of 14:58, 25 September 2008

All three arguments of the logarithm can be written as powers of \displaystyle \text{3},

\displaystyle \begin{align} & 27^{\frac{1}{3}}=\left( 3^{3} \right)^{\frac{1}{3}}=3^{3\centerdot \frac{1}{3}}=3^{1}=3, \\ & \frac{1}{9}=\frac{1}{3^{2}}=3^{-2} \\ \end{align}


and it is therefore appropriate to use base \displaystyle \text{3} when simplifying using the logarithms, even if we have the base \displaystyle \text{1}0 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} \\ & =\lg 3+\frac{1}{2}\lg 3+\left( -2 \right)\centerdot \lg 3 \\ & =\left( 1+\frac{1}{2}-2 \right)\lg 3=-\frac{1}{2}\lg 3 \\ \end{align}


This expression cannot be simplified any further.

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.3:4c"