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		<id>https://matte5.mathonline.se/index.php?action=history&amp;feed=atom&amp;title=1.5_L%C3%B6sning_4c</id>
		<title>1.5 Lösning 4c - Versionshistorik</title>
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		<updated>2026-07-23T11:35:24Z</updated>
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		<id>https://matte5.mathonline.se/index.php?title=1.5_L%C3%B6sning_4c&amp;diff=3075&amp;oldid=prev</id>
		<title>Taifun: Created page with &quot;&lt;math&gt; {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5...&quot;</title>
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				<updated>2011-03-10T12:05:54Z</updated>
		
		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;lt;math&amp;gt; {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Ny sida&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&amp;lt;math&amp;gt; {9\,^{z+1} \cdot 81\,^{3\,z/4} \over 27\,^{5\,z/3}} = {(3^2)^{z+1} \cdot (3^4)^{3\,z/4} \over (3^3)^{5\,z/3}} = {3^{2\cdot(z+1)} \cdot 3^{4\cdot(3\,z/4)} \over 3^{3\cdot(5\,z/3)}} = &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2} \cdot 3^{3\,z} \over 3^{5\,z}} = {3^{2\,z+2+3\,z} \over 3^{5\,z}} = {3^{5\,z+2} \over 3^{5\,z}} = {3^{5\,z+2-5\,z}} = 3^2 = 9 &amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>Taifun</name></author>	</entry>

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