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POWhy log(1000)/log(10) isn't the same as log10(1000)?
 

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  1. POWhy log(1000)/log(10) isn't the same as log10(1000)?
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    <p>Today, I came across quite strange problem. I needed to calculate string length of a number, so I came up with this solution</p> <pre><code>// say the number is 1000 (int)(log(1000)/log(10)) + 1 </code></pre> <p>This is based on mathematical formula</p> <p><code>log</code><sub>10</sub><code>x = log</code><sub>n</sub><code>x/log</code><sub>n</sub><code>10</code> (explained <a href="http://en.wikipedia.org/wiki/Logarithm#Change_of_base" rel="nofollow noreferrer">here</a>)</p> <p>But I found out, that in C,</p> <pre><code>(int)(log(1000)/log(10)) + 1 </code></pre> <p>is <strong>NOT</strong> equal to</p> <pre><code>(int) log10(1000) + 1 </code></pre> <p>but it should be. </p> <p>I even tried the same thing in Java with this code</p> <pre><code>(int) (Math.log(1000) / Math.log(10)) + 1 (int) Math.log10(1000) + 1 </code></pre> <p>but it behave the same wrong way.</p> <p>The story continues. After executing this code</p> <pre><code>for (int i = 10; i &lt; 10000000; i *= 10) { System.out.println(((int) (Math.log10(i)) + 1) + " " + ((int) (Math.log(i) / Math.log(10)) + 1)); } </code></pre> <p>I get </p> <pre><code>2 2 3 3 4 3 // here second method produces wrong result for 1000 5 5 6 6 7 6 // here again </code></pre> <p>So the bug seems to occur on every multiple of 1000.</p> <p>I showed this to my C teacher, and he said that it might be caused by some type conversion error during log division, but he didn't know why.</p> <p>So my questions are</p> <ul> <li><strong>Why isn't <code>(int) (Math.log(1000) / Math.log(10)) + 1</code> equal to <code>(int) Math.log10(1000) + 1 </code>, while it should be, according to the math.</strong></li> <li><strong>Why is it wrong only for multiples of 1000?</strong></li> </ul> <p><strong>edit:</strong> It is not rounding error, because </p> <pre><code>Math.floor(Math.log10(i)) + 1 Math.floor(Math.log(i) / Math.log(10)) + 1 </code></pre> <p>produce same, wrong output</p> <pre><code>2 2 3 3 4 3 5 5 6 6 7 6 </code></pre> <p><strong>edit2:</strong> I have to round down, because I want to know the <strong>number of digits</strong>.</p> <pre><code>log10(999) + 1 = 3.9995654882259823 log10(1000) + 1 = 4.0 </code></pre> <p>If I just round, I get same result (4), which is wrong for 999, because it has 3 digits.</p>
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    1. COThe unrounded output from these is: log10(1000) = 4.0 log(1000)/log(10) = 3.9999999999999996 What was the justifcation for using `floor` over using `round`?
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    2. COTry log10(999) + 1, you get 3.9995654882259823, which must be floored, because I want number of digits. If I round, I get 4, which is same as log10(1000) + 1, but it is one digit longer
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    3. COBetter use Integer.toString(n).length() to get the length.
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