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 $b .$

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