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	<title>Comments on: more users = more bugs found</title>
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	<description>you get what you pay for</description>
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		<title>By: Pete</title>
		<link>http://weblog.saardrimer.com/index.php/large-numbers-more-bugs-found_36/comment-page-1/#comment-1211</link>
		<dc:creator>Pete</dc:creator>
		<pubDate>Wed, 30 Nov 2005 16:17:02 +0000</pubDate>
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		<description>Close John.  log10(x) = ln(x)/ln(10).  Division, not multiplication.  So logb(x) = ln(x)/ln(b).</description>
		<content:encoded><![CDATA[<p>Close John.  log10(x) = ln(x)/ln(10).  Division, not multiplication.  So logb(x) = ln(x)/ln(b).</p>
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		<title>By: John</title>
		<link>http://weblog.saardrimer.com/index.php/large-numbers-more-bugs-found_36/comment-page-1/#comment-1003</link>
		<dc:creator>John</dc:creator>
		<pubDate>Sat, 19 Nov 2005 03:20:03 +0000</pubDate>
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		<description>FYI, log10(x) = ln(10)*ln(x), and, in general, the base b logarithm of a number x is equal to the natural logarithm of b times the natural logarithm of x, so any logarithm is the same as any other, up to a constant multiplier. Also, when you say that the number of bugs discovered is &quot;directly proportional&quot; to the number of users, you are stating that NumberOfBugs = NumberOfUsers*ConstantOfProportionality. What you mean to say is, &quot;the number of bugs discovered is proportional to the logarithm of the number of users&quot;.</description>
		<content:encoded><![CDATA[<p>FYI, log10(x) = ln(10)*ln(x), and, in general, the base b logarithm of a number x is equal to the natural logarithm of b times the natural logarithm of x, so any logarithm is the same as any other, up to a constant multiplier. Also, when you say that the number of bugs discovered is &#8220;directly proportional&#8221; to the number of users, you are stating that NumberOfBugs = NumberOfUsers*ConstantOfProportionality. What you mean to say is, &#8220;the number of bugs discovered is proportional to the logarithm of the number of users&#8221;.</p>
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