http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php?action=history&feed=atom&title=Answer_5.1%3A6 Answer 5.1:6 - Revision history 2026-04-08T00:50:15Z Revision history for this page on the wiki MediaWiki 1.11.1 http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php?title=Answer_5.1:6&diff=6284&oldid=prev Tek: New page: {| width="100%" cellspacing="10px" |a) |width="100%" | \ln(4\times 3)=\ln 4+\ln 3 |- |b) |width="100%" | \ln(4-3)\ne \ln 4-\ln 3 |- |c) |width="100%" | \log_{2}4 = \dfrac{\ln 4}{\ln 2} |- ... 2009-01-28T14:26:09Z <p>New page: {| width=&quot;100%&quot; cellspacing=&quot;10px&quot; |a) |width=&quot;100%&quot; | \ln(4\times 3)=\ln 4+\ln 3 |- |b) |width=&quot;100%&quot; | \ln(4-3)\ne \ln 4-\ln 3 |- |c) |width=&quot;100%&quot; | \log_{2}4 = \dfrac{\ln 4}{\ln 2} |- ...</p> <p><b>New page</b></p><div>{| width=&quot;100%&quot; cellspacing=&quot;10px&quot;<br /> |a)<br /> |width=&quot;100%&quot; | \ln(4\times 3)=\ln 4+\ln 3<br /> |-<br /> |b)<br /> |width=&quot;100%&quot; | \ln(4-3)\ne \ln 4-\ln 3<br /> |-<br /> |c)<br /> |width=&quot;100%&quot; | \log_{2}4 = \dfrac{\ln 4}{\ln 2}<br /> |-<br /> |d)<br /> |width=&quot;100%&quot; | 2^{\log_{2}4} = 4<br /> |}</div> Tek