# 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.