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 \(b.\)

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