Ask Uncle Colin: A Log Approximation

Dear Uncle Colin,

According to John D Cook, you can estimate the natural logarithm of a number $x$ by working out ${\log }_2(x)-{\log }_{10}(x)$. It seems to be pretty accurate - but why?

Not A Problem - Interesting Exponential Riddle

Hi, Napier, and thanks for your message!

That is indeed a conundrum: why should $\ln (x)\approx {\log }_2(x)-{\log }_{10}(x)$?

The simplest approach I can see is to apply the change of base formula: ${\log }_a(x)=\frac{{\log }_b(x)}{{\log }_b(a)}$. In particular, ${\log }_2(x)=\frac{\ln (x)}{\ln (2)}$ and ${\log }_{10}(x)=\frac{\ln (x)}{\ln (10)}$.

This gives us a right-hand side of $\ln (x)(\frac{1}{\ln (2)}-\frac{1}{\ln (10)})$. If that bracketed factor is about 1, then the approximation is good.

And, it turns out, it is good: $\frac{1}{\ln (2)}-\frac{1}{\ln (10)}=\frac{\ln (10)-\ln (2)}{\ln (10)\ln (2)}=\frac{\ln (5)}{\ln (2)\ln (10)}$, which we can rewrite as $\frac{{\log }_{10}(5)}{\ln (2)}$ - because those are numbers you cover in week one of Ninja school ¹

• $\ln (2)\approx 0.69315$
• ${\log }_{10}(5)=1-{\log }_{10}(2)\approx 1-0.30103=0.69897$

Those numbers are not identical, but they’re pretty close together: the difference is about six parts in 700, slightly under 1%.

Hope that helps!

- Uncle Colin

* Thanks to @pozorvlak for pointing me at John’s article.

Footnotes:

1. Or so I’m told.