Solution 1.3:4c

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
(Ny sida: {{NAVCONTENT_START}} <center> Bild:1_3_4c.gif </center> {{NAVCONTENT_STOP}})
Current revision (14:05, 22 September 2008) (edit) (undo)
m
 
(3 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
The whole expression consists of factors having a base of 5 so the power rules can be used to simplify the expression first
-
<center> [[Bild:1_3_4c.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>\begin{align}
 +
\frac{5^{12}}{5^{-4}}\cdot \bigl( 5^{2} \bigr)^{-6}
 +
&= \frac{5^{12}}{5^{-4}}\cdot 5^{2\cdot (-6)}\\[3pt]
 +
&= \frac{5^{12}}{5^{-4}}\cdot 5^{-12}\\[3pt]
 +
&= \frac{5^{12}\cdot 5^{-12}}{5^{-4}}\\[3pt]
 +
&= \frac{5^{12-12}}{5^{-4}}\\[3pt]
 +
&= \frac{5^{0}}{5^{-4}}\\[3pt]
 +
&= 5^{0-(-4)}\\[3pt]
 +
&= 5^{4}\\[3pt]
 +
&= 5\cdot 5\cdot 5\cdot 5\\[3pt]
 +
&= 625\,\textrm{.}
 +
\end{align}</math>}}

Current revision

The whole expression consists of factors having a base of 5 so the power rules can be used to simplify the expression first

\displaystyle \begin{align}

\frac{5^{12}}{5^{-4}}\cdot \bigl( 5^{2} \bigr)^{-6} &= \frac{5^{12}}{5^{-4}}\cdot 5^{2\cdot (-6)}\\[3pt] &= \frac{5^{12}}{5^{-4}}\cdot 5^{-12}\\[3pt] &= \frac{5^{12}\cdot 5^{-12}}{5^{-4}}\\[3pt] &= \frac{5^{12-12}}{5^{-4}}\\[3pt] &= \frac{5^{0}}{5^{-4}}\\[3pt] &= 5^{0-(-4)}\\[3pt] &= 5^{4}\\[3pt] &= 5\cdot 5\cdot 5\cdot 5\\[3pt] &= 625\,\textrm{.} \end{align}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_1.3:4c"