# Why is the decimal expansion of log2(3) infinite?

\(\log_2(3)\) starts with

1.5849625007211561814537389439478165087598144076924810604557526545410982277943585625222804749180882420909806624750591673437175524410609248221420839506216982994936575922385852344415825363027476853069780516875995544737266834624612364248850047581810676961316404807130823233281262445248670633898014837234235783662478390118977006466312634223363341821270106098049177472541357330110499026268818251703576994712157113638912494135752192998699040767081539505404488360

and goes on indefinitely. Why is it 1.58… in particular? Well, it takes more than one but less than two binary digits to encode a 3-digit, so \(\log_2(3)\) must be between 1 and 2. (Wait, what?). It takes more than 15 but less than 16 binary digits to encode ten 3-digits, so \(10 \cdot \log_2(3)\) must be between 15 and 16, which means \(1.5 < \log_2(3) < 1.6.\) It takes more than 158 but less than 159 binary digits to encode a hundred 3-digits, so \(1.58 < \log_2(3) < 1.59.\) And so on. Because no power of 3 is ever equal to any power of 2, \(10^n \cdot \log_2(3)\) will never quite be a whole number, no matter how large \(n\) is.

Thus, \(\log_2(3)\) has no finite decimal expansion, because \(3\) is not a rational power of \(2\). Using this argument, we can see that \(\log_b(x)\) is an integer if (and only if) \(x\) is a power of \(b\), and that \(\log_b(x)\) only has a finite expansion if some power of \(x\) is a power of \(b.\)

Parents:

\(8\) is not a power of \(4\), but \(\log_4 8\) is \(1.5\). The only thing you prove with \(3\) is not a power of \(2\) is that \(log_2 3\) is not an

integer.Fixed, thanks.