Why is the decimal expansion of log2(3) infinite?
\(\log_2(3)\) starts with
and goes on indefinitely. Why is it 1.58… in particular? Well, it takes more than one but less than two 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.to encode a
Thus, \(\log_2(3)\) has no finite decimal expansion, because \(3\) is not a rational 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 \(b.\)