Taylor Expansion of $\log \left( 1+\frac{4.645}{\sqrt{R\cdot S}} \right)$

Question

I've come across the following Taylor expansion in an old hydraulics paper $$ \log \left( 1+\frac{4.645}{\sqrt{R\cdot S}} \right)=\frac{1}{2.303}\left( x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}\ldots etc. \right) $$ where x is $$ x=\frac{4.645}{\sqrt{R\cdot S}} $$ I can see that the Taylor Expansion of $\log \left( 1+x \right)=\left( x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}\ldots etc. \right)$ so where does the $\frac{1}{2.303}$ come from. I can see it is $\frac{1}{\ln \left( 10 \right)}=\frac{1}{2.303}$, but why? Thanks in advance